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Section 4.10 Kinematics Bootcamp

Subsection 4.10.1 Position and Displacement Vectors

Subsection 4.10.2 Velocity and Speed

Subsection 4.10.3 Acceleration

Subsection 4.10.4 Displacement from Velocity

Subsection 4.10.5 Velocity from Acceleration

Subsection 4.10.6 Constant Acceleration and Projectile Motions

Subsection 4.10.7 Variable Acceleration

Subsection 4.10.8 Relative Motion

Subsection 4.10.9 Miscellaneous

Suppose you live on the \(5^{\text{th}}\) floor of a tall building. Your friend goes to the roof and drops a steel ball from rest while you time the ball as it falls. You find that the ball takes \(0.11\ \text{sec}\) to fall from the top of the window to the bottom of the window, a distance of \(0.8\) meter. How high above the top of the window of the \(5^{\text{th}}\) floor is the roof?

Hint

First find the speed at the top of the window. Then, use that as initial speed for the problem for just the window.

Answer

\(2.31\text{ m}\text{.}\)

Solution

Let \(h\) be the height from top of the window to where the ball was let go from rest. We know that after freely falling this height, the ball would be moving at \(\sqrt{2gh}\text{.}\)

Now, let us look at motion from top of the window to bottom of the window. Let \(t=0\) at the top of window and \(t=t\) at tje bottom.

Then, using positive \(y\) axis pointed up, we have the following

\begin{equation*} v_{yi} = - \sqrt{2gh},\ \ a_y = -g,\ \ \Delta t= t, \ \ y_f-y_i = -h_w. \end{equation*}

Therefore we have the following equation of constant accleration useful in this context.

\begin{equation*} y_f - y_i = v_{yi}\Delta t + \dfrac{1}{2}a_y (\Delta t)^2. \end{equation*}

This gives

\begin{equation*} -h_w = -(\,\sqrt{2gh}\,)\, t -\dfrac{1}{2} g t^2. \end{equation*}

Solving for \(\sqrt{2gh}\) we get

\begin{align*} \sqrt{2gh}\amp = \dfrac{h_w}{t} - \dfrac{1}{2} g t,\\ \amp = \dfrac{0.8}{0.11} - \dfrac{1}{2} \times 9.81\times 0.11 = 7.27 - 0.54 = 6.73. \end{align*}

Therefore,

\begin{equation*} h = \dfrac{6.73^2}{2\times 9.81} = 2.31\text{ m}. \end{equation*}

A football leaves the kicker's foot at \(50^{\circ}\) from the horizontal direction. The football is required to clear a horizontal bar at a height of \(10\text{ m}\) above the ground at a horizontal distance of \(40\text{ m}\text{.}\) Find speed with which ball must leave the kicker's foot so that it would barely go over the bar.

Hint

Standard projectile motion setup.

Answer

\(22.5\text{ m/s}\text{.}\)

Solution

Let \(h\) denote the height to clear and \(D\) the horizontal distance to travel from the launch site as shown in Figure 4.10.49. Let \(v_i\) denote the speed at the initial instant and \(\theta\) angle of launch.

Figure 4.10.49. Setup up for Problem 4.10.48.

With \(t\) the duration of the flight we have the following two equations, one along \(x\) axis with \(a_x=0\) and one along \(y\) with \(a_y=-g\) that is useful to solve this problem.

\begin{align} D \amp = v_i\, t\, \cos\,\theta,\label{eq-football-clearing-goal-x}\tag{4.10.1}\\ h \amp = v_i\, t\, \sin\,\theta -\dfrac{1}{2}g t^2,\label{eq-football-clearing-goal-y}\tag{4.10.2} \end{align}

replace \(v_i t\) in the second equation to obtain an equation in \(t\) that we can solve for \(t\text{.}\)

\begin{equation*} h = D \tan\,\theta - \dfrac{1}{2}g t^2, \end{equation*}

which gives

\begin{equation*} t^2 = \dfrac{2}{g}(D \tan\,\theta - h ) = (2/9.81)\times(40\tan\,50^\circ - 10 ) = 7.68. \end{equation*}

Therefore,

\begin{equation*} t = \sqrt{7.68} = 2.77\text{ s}. \end{equation*}

Using this in Eq. (4.10.1) we get speed at initial instant to be

\begin{equation*} v_i = \dfrac{D}{t \cos\,\theta} = \dfrac{40}{2.77\, \cos\,50^\circ} = 22.5\text{ m/s}. \end{equation*}

A rocket is rising at a constant velocity of \(200\ \text{m/s}\text{.}\) A piece of the rocket comes loose and falls freely to the ground. It takes \(40\ \text{sec}\) for the piece to fall to the ground after dislodging from the rocket. How far above the ground the piece came loose?

Hint

Use postive \(y\) axis pointed up and use correct signs for directions.

Answer

\(152\text{ m}\text{.}\)

Solution

Let positive \(y\) axis point up. Let \(h\) be the required height. Then, we have the following informaion of the free motion during the interval \(\Delta t=50\text {sec}\text{.}\)

\begin{align*} \amp y_i = h, \ y_f=0,\ v_{iy} = +200\text{ m/s},\ v_{fy} =?,\\ \amp a_y=-9.81\text{ m/s}^2,\ \ t = 40\text{ s}. \end{align*}

Using the displacement-time equation we get

\begin{equation*} 0-h =200\times 40 - \dfrac{1}{2}\times 9.81 \times 40^2. \end{equation*}

Solving for \(h\) we get

\begin{equation*} h = 152\text{ m}. \end{equation*}

One way to measure the value of \(g\) is to launch a projectile vertically upward and record the return times at two different heights. Let \(T_1\) and \(T_2\) be the two return times, i.e.,the time for 1-3-1' and 2-3-2', at heights \(H_1\) and \(H_2\) respectively as shown in Fig. \ref{fig:prob-kin-8}. Deduce the following formula for \(g\) from the heights and times.

\begin{equation*} g = \frac{8\left( H_2 - H_1\right)}{T_1^2 - T_2^2}. \end{equation*}

Hint

With same launch find the initial speeds at the two heights. Then, use that to get hang times for each height.

Answer

Ans. already in statement.

Solution

Let \(v_0\) be the launch speed, and \(v_1\) and \(v_2\) be speeds at heights \(H_1\) and \(H_2\text{.}\) Therefore we will have

\begin{align} v_1^2 \amp = v_0^2 - 2 g H_1 \label{eq-prob-kin-8-speed-height-1}\tag{4.10.3}\\ v_1^2 \amp = v_0^2 - 2 g H_1 \label{eq-prob-kin-8-speed-height-2}\tag{4.10.4} \end{align}

Now, when you launch a ball up with a speed \(v\) and if the return time is \(t\text{,}\) then you can set up the \(y_f-y_i=0\) for this to show that.

\begin{equation*} 0 = v t - \dfrac{1}{2} g t^2. \end{equation*}

That is \(gt\) of return time is

\begin{equation*} gt = 2 v. \end{equation*}

Applying this to the two return times we have

\begin{align} \amp gT_1 = 2 v_1,\tag{4.10.5}\\ \amp gT_2 = 2 v_2.\tag{4.10.6} \end{align}

Square these and subtract to get

\begin{equation*} g^2 \left( T_1^2 - T_2^2 \right) = 4\left( v_1^2 - v_2^2\right). \end{equation*}

Now using \(v_1^2\) and \(v_2^2\) from Eqs. (4.10.3) and (4.10.4), we will get

\begin{equation*} g^2 \left( T_1^2 - T_2^2 \right) = 8\, g\, \left(H_2 - H_1\right). \end{equation*}

Therefore,

\begin{equation*} g = \dfrac{ 8\, \left(H_2 - H_1\right) }{ T_1^2 - T_2^2 }. \end{equation*}