More Projectile Motion Graph Examples

A ball that always bounces off the floor with the same magnitude of the velocity as it hit the floor is dropped from \(\text{3}\) \(\text{m}\) above a stack of 4 crates. Each crate is \(\text{30}\) \(\text{cm}\) high. After the ball bounces off a crate it is quickly removed so that the next time the ball bounces it bounces off the next crate in the pile. This is repeated until all the crates have been removed. Draw the following graphs for the situation:

• Displacement vs time
• Velocity vs time
• Acceleration vs time

Solution

The time in each case for the ball to fall from maximum to the crate can be calculated from the distance:\begin{align*}s&=ut+\frac{1}{2}gt^2 \\&(u=0) \\s&=\frac{1}{2}gt^2 \\\frac{2s}{g}&=t^2 \\t&= \sqrt{\frac{2s}{g}}\end{align*}The ball bounces perfectly which means it always bounces back to the original height. The distances for each bounce are then:

• \(x_1 = \text{3}\text{ m}\)
• \(x_2 = \text{3.3}\text{ m}\)
• \(x_3 =\text{3.6}\text{ m}\)
• \(x_4 =\text{3.9}\text{ m}\)

The times to drop these distances under gravitational acceleration from rest are then:

• \(x_1= \text{3}\text{ m}~ ;~ t_1=\text{0.782460}\text{ s}\)
• \(x_2 = \text{3.3}\text{ m}~ ;~ t_2=\text{0.820651}\text{ s}\)
• \(x_3 =\text{3.6}\text{ m}~ ;~ t_3=\text{0.857142}\text{ s}\)
• \(x_4 =\text{3.9}\text{ m}~ ;~ t_4=\text{0.892142}\text{ s}\)

• \(x_1= \text{3}\text{ m}~ ;~ t_1=\text{0.782460}\text{ s}\)
• \(x_2 = \text{3.3}\text{ m}~ ;~ t_2=\text{0.820651}\text{ s}\)
• \(x_3 =\text{3.6}\text{ m}~ ;~ t_3=\text{0.857142}\text{ s}\)
• \(x_4 =\text{3.9}\text{ m}~ ;~ t_4=\text{0.892142}\text{ s}\)

A ball is dropped from \(\text{4}\) \(\text{m}\) above a cushioned mat. Each time the ball bounces the magnitude of the velocity in the upwards direction is half the magnitude of the velocity with which it hit the floor. The ball is allowed to bounce 2 times, draw the following graphs for the situation and the equation describing each section of the graph:

• Displacement vs time
• Velocity vs time
• Acceleration vs time

Solution

The time in each case for the ball to fall from maximum to the mat can be calculated from the distance:

\begin{align*}s&=ut+\frac{1}{2}gt^2 \\&(u=0) \\s&=\frac{1}{2}gt^2 \\\frac{2s}{g}&=t^2 \\t&= \sqrt{\frac{2s}{g}}\end{align*}

We calculate the velocity with which it hits the mat from:

\[v^2 =u^2+2gs\]

After the bounce it has an initial velocity upwards of 1/2 the magnitude that it hit with. Weuse this reduced initial velocity to calculate the new height.

\[v^2 =u^2+2gs\]

1. First motion:

   \begin{align*}v_{1f}^2&=2gs \\v_{1f}^2&=2(9,8)(4) \\v_{1f}&=\sqrt{2(9,8)(4)} \\v_{f1}&= \text{8,854377}\text{ m·s$^{-1}$}\end{align*}

   Time to fall from max:

   \begin{align*}s_1&=ut+\frac{1}{2}gt_1^2 \\&(u=0) \\s_1&=\frac{1}{2}gt_1^2 \\\frac{2s_2}{g}&=t_1^2 \\t_1&= \sqrt{\frac{2s_1}{g}} \\t_1&= \sqrt{\frac{2(4)}{9,8}} \\&= \text{0,9035079}\text{ s}\end{align*}

2. First bounce, height attained:

   \begin{align*}v_{max}^2&=v_{i2}^2+2gs \\&(v_{max}=0) \\-(0.5*8.854377)^2&=2(-9,8)s \\s_2&=\frac{-(0.5*8.854377)^2}{2(-9,8)} \\s_2&= \text{1,00000}\text{ m}\end{align*}

   Time to fall from max:

   \begin{align*}s_2&=\frac{1}{2}gt_2^2 \\&(v_i=0) \\s_2&=\frac{1}{2}gt_2^2 \\\frac{2s_2}{g}&=t_2^2 \\t_2&= \sqrt{\frac{2s_2}{g}} \\t_2&= \sqrt{\frac{2(1)}{9,8}} \\&= \text{0,45175395}\text{ s}\end{align*}

3. Second bounce, height attained:

   \begin{align*}v_{max}^2&=v_{i3}^2+2gs \\&(v_{max}=0) \\-(0.5*0.5*8.854377)^2&=2(-9,8)s_3 \\s_3&=\frac{-(0.5*0.5*8.854377)^2}{2(-9,8)} \\s_3&=\text{0,25000}\text{ m}\end{align*}

   Time to fall from max:

   \begin{align*}s_3&=\frac{1}{2}gt_3^2 \\&(u=0) \\s_3&=\frac{1}{2}gt_3^2 \\\frac{2s_3}{g}&=t_3^2 \\t_3&= \sqrt{\frac{2s_3}{g}} \\t_3&= \sqrt{\frac{2(0.25000)}{9,8}} \\&= 0.225877\end{align*}