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## Section4.7Constant Acceleration

Constant acceleration motion plays an important role in physics. Even when acceleration is not constant, treating the motion by using average acceleration can at times give us a rough idea of physical possibilities in the situation.

In this section, we will study standard examples of constant acceleration such as free flights of projectiles near the surface of the Earth (or any other planet) and boxes sliding on inclined planes. But before you go through the numerical exercises, it will be helpful to get a broader picture presented below.

Constant acceleration means both magnitude and direction are unchanging during the interval of interest. If acceleration changes at any instant, we must stop our analysis by constant-accceleration methods an instant before that happens.

The direction of acceleration is a special direction in this setting. Therefore, it is common practice to point one of the Cartesian axes, say $x$ axis in the direction of the acceleration, as we will see below.

### Subsection4.7.1The Plane of Constant Acceleration Motion

Choosing axes in a constant-acceleration problem is an important step. The logical steps are outlined in Figure 4.7.1.

You can always have the entire motion in the $xy$-plane by choosing your $x$ and $y$ axes appropriately. This is possible due to the fact that if your $v_z = 0$ and $a_z = 0 \text{,}$ then $v_z$ will remain zero throughout, and the object will be confined to the $xy$-plane.

To get a coordinate system in which both $v_z = 0$ and $a_z = 0 \text{,}$ we will choose the $z$-axis to be perpendicular to both the initial velocity and acceleration vectors. This will ensure that the motion will occur in the $xy$-plane.

To simplify algebra further, we will choose either the $x$ axis or the $y$ axis in the direction of the acceleration vector. This simplifies $\vec a$ to have only one non-zero component. That means the motion in the other axis will be a simple constant velocity motion.

Even more simplification can occur if we choose the origin of the coordinate system to be the location of the object at the initial instant of interest, making

\begin{equation*} x_{i} = 0,\ \ y_{i} = 0. \end{equation*}

### Subsection4.7.2Constant Acceleration Along $x$ Axis

Suppose an object, say a baseball, is moving at constant acceleration of magnitude $a$ during $t=t_i$ and $t=t_f\text{.}$ We wish to address the following question.

If position and velocity at $t_i$ are $\vec r_i$ and $\vec v_i \text{,}$ what will be the position and velocity, $\vec r_f$ and $\vec v_f \text{,}$ at $t_f \text{?}$

From our discusssion above, we already, know that we can choose $x$ and $y$ axes so that $\vec a$ and $\vec v_i$ are in the $xy$-plane. By choosing $x$ to be along the acceleration and origin at the initial position, we will have the following.

\begin{align*} \amp \vec a = (a_x, 0) = (a,0)\\ \amp \vec v_i = \left(v_{i,x}, v_{i,y}\right)\\ \amp \vec r_i = (0, 0) \end{align*}

Similarly, the position and velocity at final instant have the following components.

\begin{align*} \amp \vec r_f = (x_f, y_f)\\ \amp \vec v_f = \left(v_{f,x}, v_{i,y}\right) \end{align*}

Let us denote the duration $t_i$ to $t_f$ by $t$ rather than $\Delta t$ for the sake of brevity.

\begin{equation*} \Delta t = t_f - t_i \equiv t, \end{equation*}

Similarly, we will use a simpler notation for the displacement.

\begin{align*} \amp x \equiv \Delta x = x_f - x_i = x_f,\\ \amp y \equiv \Delta y = y_f - y_i = y_f. \end{align*}

Separate $x$ and $y$ Worlds.

Now, collecting $x$ and $y$ components in separate sets along with the time duration, we get two sets of quantities.

\begin{align} \amp x,\ v_{i,x},\ v_{f,x},\ a_x=a,\ t, \tag{4.7.1}\\ \amp y,\ v_{i,y},\ v_{f,y},\ a_y=0,\ t. \tag{4.7.2} \end{align}

The $x$ quantities do not depend on the $y$ quantities, and vice-versa, except for the value of time $t$ they share. Furthermore, $x$ quantities are simply one-dimensional constant acceleration motion and $y$ quantities are constant velocity motion. We have already studied them in the context of one-dimensional motions. Therefore, we just summarize the equations we got before.

Accleration along $x$ axis with motion in $xy$ plane:

\begin{align*} x\text{-motion:}\amp \\ \amp x = v_{i,x} \, t + \dfrac{1}{2}\, a_x \ t^2\\ \amp v_{f,x} = v_{i,x} + a_x \, t\\ \amp v_{f,x}^2 = v_{i,x}^2 + 2\, a_x \, x \\ y\text{-motion:}\amp \\ \amp y = v_{i,y} \, t \\ \amp v_{f,y} = v_{i,y} \end{align*}

### Subsection4.7.3The Projectile Motion

You have studied what happens in a free fall when you toss a projectile vertically. It just goes up and then comes down with a constant acceleration of magnitude $g$ pointed in straight down direction. When you toss a projectile at an angle, it still has the same acceleration of magnitude $g$ pointed down, but the projectile also moves in the horizontal direction as shown in Figure 4.7.2. This motion is called a projectile motion.

We could choose an axis that has positive $x$ axis down so that $a_x = g$ and $a_y = 0\text{.}$ If we do that, we can just use the formulas above for this value of $a_x\text{.}$

However, traditionally, projectile motion is solved using $y$ coordinate pointed up as shown in Figure 4.7.2. This gives us $a_x = 0$ and $a_y = -g\text{,}$ completely switching the roles of $x$ and $y$ axes we used above. Additionally, we get a minus sign. Making these adjustments we get the following formulas, where $t=t_f-t_i,\ x=x_f-x_i,\ y=y_f-y_i\text{.}$

Free Fall traditional axes:

\begin{align*} x\text{-motion:}\amp \\ \amp x = v_{i,x} \, t \\ \amp v_{f,x} = v_{i,x}\\ y\text{-motion:}\amp \\ \amp y = v_{i,y} \, t - \dfrac{1}{2}\, g \ t^2\\ \amp v_{f,y} = v_{i,y} - g \, t\\ \amp v_{f,y}^2 = v_{i,y}^2 - 2\, g \, y \end{align*}

A runner starts from rest and picks up speed at a constant rate on a straight track. The acceleration has magnitude $1.5\text{ m/s}^2\text{.}$

(a) What will be his speed $5$ seconds later?

(b) How far from the starting place will he be at that time?

Hint

(a) Use $v_{f,x} = v_{i,x} + a_x \, t$

Answer

(a) $7.5\text{ m/s}\text{,}$ (b) $18.75\text{ m}\text{.}$

Solution 1 (a)

(a) This motion is on only one axis. So, let us use $x$ axis for that with the origin at the starting place, and positive $x$ axis pointed in the direction of the runner's motion, which is also the direction of his acceleration here. With this choice, we do not need the axis of the constant velocity motion. We also set $t_i = 0$ and $t_f = t\text{.}$

With this choice, we have the following values.

\begin{equation*} a_x = 1.5\text{ m/s}^2,\ \ v_{i,x} = 0, \ \ x_i = 0, \ \ t = 5\text{ s}, \end{equation*}

and we want to find out $v_{f,x}$ in this part and $x_f$ in the next part. Looking at the list of equations we have, we find the following one to give the answer directly. We omit units in calculations and put them back at the end.

\begin{equation*} v_{f,x} = v_{i,x} + a_x\, t = 0 + 1.5\times 5 = 7.5\text{ m/s}. \end{equation*}
Solution 2 (b)

(b) We don't need to start all over. Plus, we can use the result of part (a) if that is useful. We want $x_f\text{.}$ We have some choices in the equations we can use. We omit units in calculations and put them back at the end. Let's use the following equation.

\begin{equation*} x_f = x_i + v_{i,x}\, t + \frac{1}{2}\, a_x\, t^2, \end{equation*}

gives

\begin{equation*} x_f = 0 + 0 + \frac{1}{2}\times 1.5 \times 5^2 = 18.75\text{ m}. \end{equation*}

A ball is thrown straight up with a speed of $29.43\text{ m/s} \text{.}$ The ball passes a point P that is a height $15 \text{ m}$ above the launch point twice, once its way up and another time on its way down. Assume free fall, meaning, the accelertion is pointed down with magnitude $g$ or $9.81\text{ m/s}^2\text{.}$

(a) If the time at the instant of launch has value $t=0 \text{,}$ what are the times for the instants when the ball is at P? (b) What are the velocities of the ball when it is at P those two instants?

(c) How high does ball go before returning? (d) How long does it take to get to the highest point?

Hint
You have two separate problems: (a) and (b) make up one problem and (c) and (d) the other problem. Set up $y$ component equation only.
Answer

(a) $t_1 = 0.56\text{ sec}\text{,}$ $t_2 = 5.44\text{ sec} \text{,}$ (b) $23.94\text{ m/s}\text{,}$ $-23.94\text{ m/s}\text{,}$ (c) $44\text{ m}\text{,}$ (d) $3.0\text{ sec}\text{.}$

Solution 1 (a) and (b)

(a) and (b). Note that the initial and final points of parts (c) and (d) are different from those of parts (a) and (b). That means we have two separate problems. Lets work out (a) and (b), then we will work out (c) and (d).

We will label quantities with either subscript $i$ or $f$ corresponding to the initial point, i.e. when the ball has left the launcher, or the final point, i.e., when the ball is at point P.

Since only $y$ components are non-zero, we work with $y$ components of all quantities only. We are given:

\begin{align*} \amp a_y = -9.81 \text{ m/s}^2,\ \ v_{i,y} = + 29.43\ \text{ m/s} \\ \amp y_i = 0,\ \ y_f = +15.0 \text{ m},\ \ t_i = 0 \end{align*}

The following quantities are not known:

\begin{gather*} t = t_f-t_i,\ \ v_{f,y}, \end{gather*}

where $t$ will have two values, $t_1$ and $t_2\text{,}$ and $v_{f,y}$ will also have two values, one for going up and one for coming down.

Since the acceleration between $t_i$ and $t_f$ is constant, we use constant-acceleration equations for these components. The following choices are useful.

\begin{align} \amp y_f - y_i = v_{i,y} t + \dfrac{1}{2}\, a_y\, t^2\label{eq-problem-vertical-motion-yt}\tag{4.7.3}\\ \amp v_{f,y} = v_{i,y} + a_y t\label{eq-problem-vertical-motion-vt}\tag{4.7.4} \end{align}

We will solve Eq. (4.7.3) to obtain the times. Putting numerical values in this equation we find that $t$ obeys a quadratic equation, which will have two solutions, corresponding to the two times.

\begin{equation*} 15 - 0 = 29.43\, t + \dfrac{1}{2}\, (-9.81)\, t^2 \end{equation*}

This gives two answers, $t = 0.56, 5.44\text{.}$ Therefore,

\begin{equation*} \text{(a) }t_1 = 0.56\text{ sec, } t_2 = 5.44\text{ sec}. \end{equation*}

(b) Now we use these times to find the $v_{f,y}$ at those instants.

\begin{align*} v_{f,y,up} \amp = v_{i,y} + a_y t_1\\ \amp = 29.43 + (-9.81) \times 0.56 = 23.94\text{ m/s} \end{align*}
\begin{align*} v_{f,y,down} \amp = v_{i,y} + a_y t_2\\ \amp = 29.43 + (-9.81) \times 5.44 = -23.94\text{ m/s} \end{align*}

Two values of $v_y \text{?}$ Note that the only difference between up and down $v_y$ are the sign. When the ball is going up, the direction of the velocity is same as the direction of the positive $y$ axis, while when the ball is coming down, the direction is just the opposite, and hence the minus sign.

Solution 2 (c) and (d)

Note that at the highest point, $v_y = 0\text{.}$ Therefore, taking the launch point as the initial point and the highest point as the final point we have the following knowns:

\begin{align*} \amp a_y = -9.81 \text{ m/s}^2,\ \ v_{i,y} = + 29.43\ \text{ m/s} \\ \amp v_{f,y} =0,\ \ y_i = 0, \ \ t_i = 0 \end{align*}

The following quantities are not known:

\begin{gather*} t = t_f-t_i,\ \ y_{f}, \end{gather*}

where $y_f -y_i$ is the total height the ball flew. We now notice that we can get $t$ from using Eq. (4.7.4), which we can use in the other equation to get $y_f \text{.}$

\begin{equation*} 0 = 29.43 + (-9.81)\, t\ \ \longrightarrow \ \ t = 3.0. \end{equation*}

Hence, it takes $3.0\text{ sec}$ to get to the top of the trajectory. Using this value in Eq. (4.7.3) we get

\begin{equation*} y_f = 0 + 29.43 \times 3.0 + \frac{1}{2}\, (-9.81)\times 3.0^2 = 44. \end{equation*}

Therefore, the ball flies to a height of $44\text{ m}\text{.}$

Why is $v_y = 0$ at the highest point?

We notice that just before the ball reaches the highest point, the $y$ component of its velocity $(v_y)$ is positive, but immediartely after the highest point, $v_y$ is negative. Since velocity components must be continuous functions of time (otherwise acceleration will not be defined), the only way they can turn from positive number to negative number is that they go through zero. That means that at the highest point $v_y = 0$

A cannon ball is fired horizontally from a hilltop with a speed of $50\text{ m/s}\text{.}$ The ball lands in the valley after falling a height of $40\text{ m} \text{.}$ Assume free fall.

(a) How far did the ball fly in the horizontal direction?

(b) With what speed did the ball hit the ground?

(c) How long was the ball in the air?

Hint

Systematically set up $x$ and $y$ equations.

Answer

(a) $143 \text{ m} \text{,}$ (b) $57.3 \text{ m/s}\text{,}$ (c) $2.86 \text{ sec}.$

Solution

The setup:

The free fall assumption makes it a constant acceleration situation with $a = g$ pointed down. As usual, let positive $x$ be in the horizontal direction towards which the ball flies and positive $y$ vertically up. Let origin be at the starting place, i.e., at the top of the hill - that will make $y_f = -40\text{ m}\text{.}$ Let $t$ be the total duration between the initial and final instants.

The components:

\begin{align*} \amp \text{Initial position }: x_i = 0,\ y_i = 0,\\ \amp \text{Initial velocity }: v_{i,x} = 50\text{ m/s},\ v_{i,y} = 0,\\ \amp \text{Final position }: x_f = ?,\ y_i = -40\text{ m},\\ \amp \text{Final velocity }: v_{f,x} = ?,\ v_{f,y} = ?,\\ \amp \text{Constant acceleration }: a_x = 0,\ a_y = -9.81\text{ m/s}^2. \end{align*}

The $x$ and $y$ equations:

Using the values above we get the following equations to work with. We omit units in equations, but out them back at the end.

\begin{align*} \amp v_{f,x} = v_{i,x} = 50\text{ m/s},\\ \amp x_f = 50\,t ,\\ \amp v_{f,y} = -9.81\, t,\\ \amp -40 = -\dfrac{9.81}{2}\, t^2. \end{align*}

Solving the equations :

Solving for $t$ we find two values, $t = \pm 2.86 \text{,}$ of which only the positive duration is what we want - the negative value refers to a time when the cannon ball “would have been” at the same height if the trajectory of the ball was exteded back in time. We use the positive time value in other equations to get the other quantities. We now list the answer.

\begin{align*} \amp \text{(c) } t = 2.86 \text{ sec}. \\ \amp \text{(a) } x_f = 143 \text{ m}. \\ \amp v_{f,x} = 50\text{ m/s},\\ \amp v_{f,y} = -28\, \text{ m/s}. \end{align*}

To find the speed of the ball at the final instant before hitting the ground we find the magnitude of the final velocity from the components.

\begin{equation*} \text{(b) } v = \sqrt{50^2 + 28^2} = 57.3 \text{ m/s}. \end{equation*}

A cannon ball is fired at a $40^{\circ}$ above the horizontal direction from a hilltop with a speed of $50\text{ m/s}\text{.}$ The ball lands in the valley after falling a height of $40\text{ m} \text{.}$ Assume free fall.

(a) How far did the ball fly in the horizontal direction?

(b) With what speed did the ball hit the ground?

(c) How long was the ball in the air?

Hint

Systematically set up $x$ and $y$ equations.

Answer

(a) $292\text{ m} \text{,}$ (b) $57.4\text{ m/s} \text{,}$ (c) $7.62\text{ sec}\text{.}$

Solution

The setup:

I will use the same setup as in Problem Checkpoint 4.7.5. As usual, let positive $x$ be in the horizontal direction towards the ball flies and positive $y$ vertically up, which makes $a_y=-9.81\text{ m/s}^2$ and $a_x=0\text{.}$ Let origin be at the starting place, i.e., at the top of the hill - that will make $y_f = -40\text{ m}\text{.}$ Let $t$ be the total duration between the initial and final instants.

The components:

\begin{align*} \amp \text{Initial position }: x_i = 0,\ y_i = 0,\\ \amp \text{Initial velocity }: v_{i,x} = 50\, \cos\, 40^{\circ}=38.3,\ v_{i,y} = 50\, \sin\, 40^{\circ}=32.1,\\ \amp \text{Final position }: x_f = ?,\ y_i = -40\text{ m},\\ \amp \text{Final velocity }: v_{f,x} = ?,\ v_{f,y} = ?,\\ \amp \text{Constant acceleration }: a_x = 0,\ a_y = -9.81\text{ m/s}^2. \end{align*}

The $x$ and $y$ equations:

Using the values above we get the following equations to work with. We omit units in equations, but out them back at the end.

\begin{align*} \amp v_{f,x} = v_{i,x} = 38.3\text{ m/s},\\ \amp x_f = 38.3\,t ,\\ \amp v_{f,y} = 32.1-9.81\, t,\\ \amp -40 = 32.1\, t-\dfrac{9.81}{2}\, t^2. \end{align*}

Solving the quadratic equation for $t$ we get two equations, of which we drop the negative time equation.

\begin{equation*} t = \dfrac{1}{9.81}\left( 32.1 + \sqrt{32.1^2 + 2\times 9.81\times 40} \right) = 7.62\text{ sec}. \end{equation*}

Using this we get the $x_f$ and $v_{f,y}\text{.}$

\begin{align*} \amp x_f = 38.3\times 7.62 = 292\text{ m}.\\ \amp v_{f,y} = 32.1-9.81\times 7.62 = -42.7\text{ m/s}. \end{align*}

From $v_{f,x}$ and $v_{f,y}$ the speed when the ball hits the ground is

\begin{equation*} v = \sqrt{38.3^2 + 42.7^2} = 57.4\text{ m/s} \end{equation*}

Note: This time the ball flew much farther compared to the cannon shot horizontally in Checkpoint 4.7.5.

A football is kicked towards the East. The ball leaves the foot of the kicker at $1.0\text{ m}$ above ground. At the moment the ball leaves the foot it has a velocity of $22.0\text{ m/s}$ in $30^{\circ}$ above the horizontal direction. Ignore effect of air drag on the ball and assume ideal free fall. That is, you can assume that acceleration of the ball has magnitude $g$ and directed Down.

(a) Where on the ground will the ball land?

(b) What will be the be velocity with which the ball will strike the ground?

(c) How long the ball is in the air?

(d) In a coordinate system in which positive $x$ is towards East and positive $y$ is pointed Up, and the origin is at the place where the ball left the kicker, find the path of the ball by giving the function $y(x) \text{.}$

Hint

Separate $x$ and $y$ motions.

Answer

(a) $44.27\text{ m}\text{,}$ (b) $(19.0\text{ m/s},-11.9\text{ m/s} \text{,}$ or, $22.4\text{ m/s},\ 32^{\circ}$ clockwise from the forward (i.e., the horizontal) direction, (c) $2.33\text{ sec}\text{,}$ (d) $y = 0.58\, x - 0.014\, x^2.$

Solution 1 (a), (b), and (c)

(a), (b), and (c). Notice the initial instant $t_i=0$ is the immediate instant after the ball left the foot and the final instant of interest is $t_f = t\text{,}$ which is the instant immediately before the ball lands.

The initial velocity is given in the magnitude-direction form. For formulas, we need it in component form.

\begin{align*} v_{i,x} \amp = v_i\, \cos\, \theta\\ \amp = 22.0 \times \cos\, 30^{\circ} = 19.0\text{ m/s}. \\ v_{i,y} \amp = v_i\, \sin\, \theta\\ \amp = 22.0 \times \sin\, 30^{\circ} = 11.0\text{ m/s}. \end{align*}

During $t=0$ to $t=t$ interval, we organize $x$ and $y$ components separately, with the understanding that $t$ is same for both.

The x-world:

\begin{align*} \amp x_i = 0,\ \ x_f = ?,\ \ v_{i,x} = 19.0\text{ m/s},\ \ v_{f,x} = ?, \ \ a_x = 0, \end{align*}

The y-world:

\begin{align*} \amp y_i = 0,\ \ y_f = -1.0\text{ m},\ \ v_{i,y} = 11.0\text{ m/s},\ \ v_{f,y} = ?, \ \ a_y = - g. \end{align*}

Now we will write four equations which can be solved to get the four unknowns here: $t,\ x_f,\ v_{f,x},\ v_{f,y} \text{.}$ We will use the following set of equations

\begin{align*} v_{f,x} \amp= v_{i,x},\\ x_f \amp= x_i + v_{i,x}\, t,\\ v_{f,y} \amp= v_{i,y} + a_y\ t,\\ y_f \amp = y_i+ v_{i,y}\, t + \dfrac{1}{2}\, a_y\, t^2. \end{align*}

Putting numerical values (leaving out units) we get

\begin{align} v_{f,x} \amp = 19.0 ,\tag{4.7.5}\\ x_f \amp = 0 + 19.0\, t,\label{eq-cp-football-flight-xf}\tag{4.7.6}\\ v_{f,y} \amp = 11.0 - 9.81\ t,\label{eq-cp-football-flight-vfy}\tag{4.7.7}\\ -1.0 \amp = 0+11.0\, t - \dfrac{1}{2}\times 9.81 \, t^2. \tag{4.7.8} \end{align}

From this list is clear that we solve the last equation for $t$ and use appropriate value in middle two equations to get $x_f$ and $v_{f,y} \text{.}$ Note that when we solve the last equation for $t \text{,}$ we will get two values, of which we will choose the positive value since, we don't want the time before $t = 0\text{.}$ Simplifying we get

\begin{equation*} 4.905\, t^2 - 11\, t - 1 = 0. \end{equation*}

This gives

\begin{equation*} t = \dfrac{1}{9.81}\left( 11 \pm \sqrt{11^2 + 19.62} \right) = -0.1, 2.33. \end{equation*}

The value we want is the positive time.

\begin{equation*} t = 2.33\text{ sec}. \end{equation*}

Now we put this value in Eq. (4.7.6) and (4.7.7).

\begin{align*} \amp x_f = 19.0\times 2.33 = 44.27\text{ m},\\ \amp v_{f,y} = 11.0 - 9.81\times 2.33 = -11.9\text{ m/s}. \end{align*}

To summarize:

\begin{align*} \amp \text{(a) } x_f=44.27\text{ m}, \\ \amp \text{(b) }\vec v_f = (19.0\text{ m/s},-11.9\text{ m/s}),\\ \amp \text{(c) }t = 2.33\text{ sec}. \end{align*}

If interested, one can get final velocity in magnitude-direction form as usual.

\begin{gather*} v_f = \sqrt{19^2 + 11.9^2} = 22.4\text{ m/s},\\ \theta = \tan^{-1}\left(\dfrac{-11.9}{19} \right) = -32^{\circ}, \end{gather*}

which gives the direction to be $32^{\circ}$ clockwise from the positive $x$ axis in the fourth quadrant, i.e., below the horizontal direction.

Solution 2 (d)

Writing $x$ for $x_f\text{,}$ $y$ for $y_f\text{,}$ and $t$ for $t_f\text{,}$ we found the following for $x$ and $y$ of the ball.

\begin{align*} \amp x = 19 t, \\ \amp y = 11\, t - 4.905\, t^2. \end{align*}

We can solve the first equation for $t$ and use that in the second equation to get an equation relating $x$ to $y \text{.}$

\begin{equation*} y = \dfrac{11}{19}\, x - 4.905\times \left( \dfrac{1}{19} \right)^2\, x^2 \end{equation*}

Simplifying we get the equation of the trajectory to be

\begin{equation*} y = 0.58\, x - 0.014\, x^2. \end{equation*}

A projectile is launched at an angle $\theta$ above the horizontal direction with speed $v_0\text{.}$ The projectile lands at the same height after flying freely under gravity with acceleration due to gravity $g\text{.}$

(a) How far will it go in the horizontal direction? This is called range of the projectile.

(b) How long was the ball in the air? This is called the hang time.

Hint

Set up $x$ and $y$ equations.

Answer

(a) $\dfrac{v_0^2}{g}\,\sin\,2\theta\text{,}$ (b) $\dfrac{2}{g}\,v_0\,\sin\,\theta\text{.}$

Solution

We use $x$ axis horizontally and $y$ axis vertically up. Let $D$ denote the range and $T$ the hang time. The useful $x$ and $y$ equations are

\begin{align*} \amp x\text{-eq: }\ \ D = T\, v_0\,\cos\,\theta, \\ \amp y\text{-eq: }\ \ 0 = T\, v_0\,\sin\,\theta -\dfrac{g}{2}\, T^2, \end{align*}

From the $y$-eq, we get two values for $T\text{.}$

\begin{equation*} T = 0,\ \dfrac{2}{g}\,v_0\,\sin\,\theta, \end{equation*}

where $T=0$ is for the starting place and not of interest.

\begin{equation*} T = \dfrac{2}{g}\,v_0\,\sin\,\theta. \end{equation*}

Now, we use this in the $x$-eqn to get

\begin{equation*} D = \dfrac{v_0^2}{g}\,\sin\,2\theta. \end{equation*}

A cannon ball is fired up an inclined road as shown in Figure 4.7.10. The initial speed of the ball is fixed at $v_0\text{,}$ but its direction can be changed by changing $\theta$ of the the barrel. The projectile lands at a distance $D$ on the incline.

(a) Find angle $\theta$ that will give the maximum range, i.e., largest value of $D\text{.}$ (b) What is the largest $D$ that can be achieved.

Answer Hint: When $\alpha=15^\circ\text{,}$ $\theta_\text{max} = 37.5^\circ\text{.}$

Hint

Try a coordinate system with $x$-axis up the incline and $y$-axis perpendicular to that. This will give you $a_x$ and $a_y$ both non-zero. Also, keep your calculations in $v_{0x}\text{,}$ $v_{0y}\text{,}$ $a_x\text{,}$ and $a_y$ till the end to keep the calcualtion from getting cluttered with too many symbols. Last steps may have tangents and cotangents.

Solution

No Solution provided.