Tom is standing on the roof of a five-story building of height $H=5L$. Jerry is in a second-floor apartment immediately below Tom that is a height $h=2L$ above the ground. Jerry drops a ball out his window, while at the same time, Tom throws a second ball downward with a velocity that has a magnitude of $v_0$. The height of each floor of the building is $L=9.8\ m$.

Jerry drops a ball out his window and, at the same time, Tom throws a ball downward with speed $v_0$.

1. What is the smallest speed $v_{min}$ that Tom’s ball can be thrown if it hits Jerry’s ball before Jerry’s ball hits the ground?

View answer

We sketch the situation below and show the initial speeds and heights of each ball.

The kinematic equations of the position and velocity of the ball dropped by Tom are

\left\{ \begin{array}{c} \displaystyle{y_0\left(t\right)=-\frac{1}{2}gt^2-v_0t+5L} \\ \\ v_y\left(t\right)=-gt-v_0 \end{array} \right.

The kinematic equations of the position and velocity of the ball thrown down by Jerry are

\left\{ \begin{array}{c} \displaystyle{y_1\left(t\right)=-\frac{1}{2}gt^2+2L} \\ \\ v_y\left(t\right)=-gt \end{array} \right.

First, let us find the time $t_1$ at which Jerry’s ball hits the ground. This time is such that $y_1\left(t_1\right)=0$ and we solve for $t_1$ as follows

\begin{aligned} y_1\left(t_1\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_1+2L=0 \\ \\ &\Rightarrow \ \ \ \ \ \ t_1=\sqrt{\frac{4L}{g}} \end{aligned}

Second, let us find the time $t_2$ at which the two balls collide (assuming they do). This time is such that $y_0\left(t_2\right)=y_1\left(t_2\right)$ and we solve for $t_2$ as follows

\begin{aligned} y_0\left(t_2\right)=y_1\left(t_2\right)\ \ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_2-v_0t_2+5L=-\frac{1}{2}gt^2_2+2L \\ \\ &\Rightarrow \ \ \ \ \ \ t_2=\frac{3L}{v_0} \end{aligned}

Thus, the two balls collide before Jerry’s ball hits the ground if $t_2\le t_1$ which leads to the following condition on $v_0$

\begin{aligned} \frac{3L}{v_0}\le \sqrt{\frac{4L}{g}}\ \ \ \ &\Rightarrow \ \ \ \ \ \frac{v_0}{3L}\ge \sqrt{\frac{g}{4L}} \\ \\ &\Rightarrow \ \ \ \ \ v_0\ge \frac{3}{2}\sqrt{gL}=v_{min} \end{aligned}

The minimum speed $v_{min}$ with which Tom must throw his ball so that it collides with Jerry’s ball before the latter hits the ground is therefore equal to

\boxed{v_{min}=\frac{3}{2}\sqrt{gL}=\frac{3}{2}\cdot 9.8=14.7\ m/s}

2. If $v_0=2\ v_{min}$, Jerry’s ball travels a distance $d$ before the two balls collide. What is $d/h$?

View answer

If $v_0=2v_{min}=3\sqrt{gL}$ then the balls collide at date $t_2$ equal to

t_2=\frac{3L}{2v_{min}}=\frac{3L}{3\sqrt{gL}}=\sqrt{\frac{L}{g}}

At time $t_2$, the height of Jerry’s ball is equal to

y_1\left(t_2\right)=-\frac{1}{2}gt^2_2+2L=-\frac{1}{2}g{\left(\sqrt{\frac{L}{g}}\right)}^2+2L=2L-\frac{L}{2}=\frac{3L}{2}

The distance travelled by Jerry’s ball during that time is therefore equal to

d=2L-y_1\left(t_2\right)=2L-\frac{3L}{2}=\frac{L}{2}=\frac{h}{4}

where we substituted $h=2L$. The ratio $d/h$ is therefore equal to

d=\frac{h}{4} \ \ \ \ \ \Rightarrow \ \ \ \ \ \boxed{\frac{d}{h}=\frac{1}{4}}

Previous Topic

Back to Lesson

Next Lesson