Chapter 2, Problem 17STP

To determine

The time when the second ball hits the ground and the distance it travels horizontally.

Answer to Problem 17STP

The second ball hits the ground in 5 s and it would fall at a distance of 70 m from the vertical fall ball.

Explanation of Solution

Given:

The velocity of the second ball is 13 m/s horizontally.

Formula used:

The average speed is

$s=\frac{d}{t}$..... (1)

Here, $d$ and $t$ are distance and time taken to travel.

Calculation:

Rearrange equation (1) for distance,

$d=s\times t$..... (2)

The acceleration on the ball falling from a height would be the same. The time taken to fall would be the same if it is released from the same height.

The time of fall of the second ball which is released horizontally also reaches the ground in 5 seconds because the vertical force or acceleration is the same for both balls.

Substitute $13\text{m/s}$ for $s$ and 5 s for $t$ in equation (2),

  $\begin{array}{l}d=\left(13\text{m/s}\right)\left(5\text{s}\right)\\ \boxed{d=70\text{m}}\end{array}$

Conclusion:

Hence,the second ball hits the ground in 5 s and it would fall at a distance of 70 m from the first fall.