Problem:

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

Solution:

To illustrate the problem, consider the following figure:

Part A

The problem states that the initial velocity is horizontal, this means that the initial vertical velocity is zero. We are also given the height of the building (which is a downward displacement), so we can solve for the time of flight using the formula y=v_oyt+1/2at². That is,

\begin{align*}
\text{y} & =\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2 \\
 -60\:\text{m}&=0+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\text{t}^2 \\
\text{t}^2 & =\dfrac{-60\:\text{m}}{-4.905\:\text{m/s}^2} \\
\text{t}^2 & =12.2324\:\text{s}^2 \\
\text{t} & =3.50\:\text{s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

\end{align*}

Part B

To solve for the v_ox, we will use the formula \text{v}_{\text{ox}}=\frac{\Delta \:\text{x}}{\text{t}}.

\begin{align*}
\text{v}_{\text{ox}} & =\frac{100\:\text{m}}{3.50\:\text{s}} \\
\text{v}_{\text{ox}} & =28.6\:\text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part C

To solve for the velocity as the ball hits the ground, we shall consider two points: (1) at the beginning of the flight, and (2) when the ball hits the ground.

We know that the initial velocity, v_oy, is zero. To solve for the final velocity, we will use the formula \text{v}_{\text{f}}=\text{v}_{\text{o}}+\text{at}.

\begin{align*}
\text{v}_{\text{f}} & =0+\left(-9.81\:\text{m/s}^2\right)\left(3.50\:\text{s}\right) \\
\text{v}_{\text{f}} & =-34.3\:\text{m/s}

\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The negative velocity indicates that the motion is downward.

Part D

Since we already know the horizontal and vertical components of the velocity when it hits the ground, we can find the resultant.

\begin{align*}
\text{v} & =\sqrt{\text{v}_{\text{x}}^2+\text{v}_{\text{y}}^2} \\
\text{v} & =\sqrt{\left(28.57\:\text{m/s}\right)^2+\left(-34.34\:\text{m/s}\right)^2} \\
\text{v} & =44.7\:\text{m/s}

\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The direction of the velocity is

\begin{align*}
\theta_{\text{x}} & =\tan ^{-1}\left|\frac{\text{v}_{\text{y}}}{\text{v}_{\text{x}}}\right| \\
\theta _{\text{x}} & =\tan ^{-1}\left|\frac{-34.34}{28.57}\right| \\
\theta _{\text{x}} & =50.2^{\circ}

\ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

The velocity is directed 50.2° down the x-axis.