P01-110 – Application: Free-Fall

Definition: free-fall

An object is said to undergo free-fall if it moves through the atmosphere subjected solely to its weight.

Consequence:

Its acceleration is then directed toward the center of the earth and is equal to the acceleration due to gravity $g=9.8\ m/s^2$ (in magnitude).

Note: this means that all other forces (such as air resistance, drag, thrust, etc.) are neglected.

Free-Fall from Rest

A rock (point mass) is released from rest from a height $H$ above the ground as shown below. In order to illustrate an application of kinematics, we will find the time it takes the rock to reach the ground as well as its velocity when it reaches the ground.

Its acceleration is $a_y=-g$, its initial velocity is $v_{0y}=0$, and its initial height is $y_0=H$. The kinematic equations that describe the free-fall of the rock are therefore

\begin{aligned}
y\left(t\right)&=-\frac{1}{2}gt^2+H \\
v_y\left(t\right)&=-gt
\end{aligned}

Time to reach the ground:

The rock reaches the ground at a time $t_f$ such that $y\left(t_f\right)=0$ i.e.

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

Thus, the time it takes the rock to reach the ground is equal to

\boxed{t_f=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\cdot 19.6}{9.8}}=2\ s}

Velocity upon impact:

The velocity of the rock when it hits the ground is equal to

\boxed{v_y\left(t_f\right)=-gt_f=-9.8\cdot 2=-19.6\ \ m/s}

where the negative sign indicates that the velocity is directed downward.

Note: a common misconception is to think that the velocity upon impact is zero because we see the rock coming to a stop. The rock’s velocity is zero after it hits the ground, but upon impact its velocity is nonzero (the ground slows it down and brings it to a stop).

Free-Fall with Initial Velocity

Its acceleration is $a_y=-g$, its initial velocity is $v_{0y}=-v_0$, and its initial height is $y_0=H$. The kinematic equations that describe the free-fall of the rock are therefore

\begin{aligned}
y\left(t\right)&=-\frac{1}{2}gt^2-v_0t+H \\
v_y\left(t\right)&=-gt-v_0
\end{aligned}

Time to reach the ground:

The rock reaches the ground at a time $t_f$ such that $y\left(t_f\right)=0$ i.e.

\begin{aligned}
y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_f-v_0t_f+H=0 \\
\\
&\Rightarrow \ \ \ \ \ \ gt^2_f+2v_0t_f-2H=0 \\
\\
&\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c}
\displaystyle{t_f=\frac{-2v_0-\sqrt{4v^2_0+8gH}}{2g}\ <\ 0} \\ 
\\
\displaystyle{t_f=\frac{-2v_0+\sqrt{4v^2_0+8gH}}{2g}\ >\ 0} \end{array}
\right.
\end{aligned}

Ruling out the negative root, we conclude that the rock hits the ground at time

\boxed{t_f=\frac{-2v_0+\sqrt{4v^2_0+8gH}}{2g}=\frac{-2\cdot 4.9+\sqrt{4\cdot {49}^2+8\cdot 10\cdot 19.6}}{2\cdot 10}\approx 1.56\ \ s}

Velocity upon impact:

The velocity of the rock when it hits the ground is equal to

\boxed{v_y\left(t_f\right)=-gt_f-v_0=-9.8\cdot 1.56-4.9=-20.18\ \ m/s}

where the negative sign indicates that the velocity is directed downward.

Note: a common misconception is to think that the velocity upon impact is zero because we see the rock coming to a stop. The rock’s velocity is zero after it hits the ground, but upon impact its velocity is nonzero (the ground slows it down and brings it to a stop).