A ball is thrown horizontally from the roof of a building 7.5 m tall and lands 9.5 m from the base. What was the ball’s initial speed?

A ball is being thrown horizontally from a building’s roof. So, the ball moves downward with some acceleration.

The ball moves downward under the effect of gravity. It experiences a gravitational force in the downward direction due to the pull exerted by gravity.

The given data can be listed below as:

• The height of the building is $y=7.5\mathrm{m}$.
• The horizontal distance covered by the ball is $x=9.5\mathrm{m}$.
• The acceleration experienced by the ball in the downward direction is $a=g=9.81\mathrm{m}/{\mathrm{s}}^2$.

The horizontal velocity of the ball can be expressed as:

$v_x=\frac{x}{t}\dots \left(i\right)$

Here, t is the time taken by the ball.

The vertical distance or height of the building can be expressed as:

$\begin{array}{c}y=v_{i}t+\frac{1}{2}a{t}^2\\ =v_{i}t+\frac{1}{2}g{t}^2\end{array}$

Here, $v_i$is the initial velocity of the ball in the vertical direction, and g is the acceleration due to gravity whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$.

Substitute the values as 0 m/s for $v_i$, 7.5 m for y, and $9.81\mathrm{m}/{\mathrm{s}}^2$for g in the above equation.

$\begin{array}{c}7.5\mathrm{m}=\left(0\mathrm{m}/\mathrm{s}\right)\times t+\frac{1}{2}\times 9.81\mathrm{m}/{\mathrm{s}}^2\times t^2\\ t^2=1.529{\mathrm{s}}^2\\ t=1.236\mathrm{s}\end{array}$

Substitute the values as $1.236\mathrm{s}$for t, and $9.5\mathrm{m}$for x in equation (i).

$\begin{array}{c}{v}_x=\frac{9.5\mathrm{m}}{1.236\mathrm{s}}\\ =7.686\mathrm{m}/\mathrm{s}\end{array}$

Thus, the ball’s initial velocity is $7.686\mathrm{m}/\mathrm{s}$.