P01R2015 – Vertical Free-fall

This problem deals with the simplest case of motion with a constant, non-zero acceleration which is letting an object fall in the atmosphere. It is then said that the object is undergoing free fall, meaning its acceleration is constant and is equal to that due to gravity (noted $g=9.8\ m/s^2$). To find how long it takes the ball to hit the ground we need to setup the kinematic equations that will describe this problem. Practically speaking, we need to identify $a_y$, $v_{0y}$ and $y_0$ to use the following equations

\left\{ \begin{array}{c}
v_y\left(t\right)=v_{0y}+a_yt \\
\\
y\left(t\right)=y_0+v_{0y}t+\frac{1}{2}a_yt^2 \end{array}
\right.

Here, if you choose your origin to be at ground level, you will have $y_0=H$ m. The problem states that the object is initially at rest and therefore we conclude that $v_{0y}=0\ m/s$. The only acceleration that the object is subject to is $\overrightarrow{g}$ and therefore $a_y=-g=-9.8\ m/s^2$ (because $\overrightarrow{g}$ points down and up is positive here). Therefore, the kinematic equations that describe this problem are:

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

To find the time $t_1$ it takes the ball to hit the ground, you need to tell your equations that the ball hits the ground i.e. that its height it then zero. Thus, the time $t_1$ is given by

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

Therefore, we conclude that the ball strikes the ground at time $t_1$ equal to

\boxed{t_1=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\cdot 100}{9.8}}\approx 4.51\ s}

The velocity of the ball when it hits the ground it given by the value of $v_y\left(t\right)$ at time $t_1$ and is equal to

\boxed{v_y\left(t_1\right)=-gt_1=-g\sqrt{\frac{2H}{g}}=-\sqrt{2gH}=-\sqrt{2\cdot 9.8\cdot 100}\approx -44.27\ m/s}

Note that the velocity is negative because the velocity vector points downward when the ball strikes the ground.

If the ball bounces off the floor with the same speed as the impact speed, then it will go back up to its initial height $H$ because it will take as long for gravity to decelerate the ball from $v_{impact}$ to $0$ as it did to accelerate it from $0$ to $v_{impact}$.

This can be quickly derived by using the kinematic equation $v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y$ between the ground and the height $h$ where the ball turns around (stops moving upward) after bouncing off the ground. Recalling that $a_y=-g$ and that $v_{fy}=0$, we solve for $h$ as follows

\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ 0=v^2_{0y}-2g\left(y_f-y_0\right) \\
\\
&\Rightarrow \ \ \ \ \ 0=v^2_{0y}-2g\left(h-0\right)\\
\\
&\Rightarrow \ \ \ \ \ h=\frac{v^2_{0y}}{2g} \\
\\
&\Rightarrow \ \ \ \ \ \boxed{h=\frac{{\left(-\sqrt{2gH}\right)}^2}{2g}=\frac{2gH}{2g}=H}
\end{aligned}