# Question c5788

Jun 9, 2015

The distances are 35 m between the first two balls, 25 m between the second and the third, and 15 m between the third and the fourth.

#### Explanation:

Here's how you should approach this problem.

You know that the first ball you drop reaches the ground after 4 s of free fall. The fact that they are released at 1 second intervals after the first ball implies that the other three balls are still in the air whn the first one touches the ground.

The first thing you need to determine is the height from which the balls were dropped. The equation that you'll use looks like this

$h = {v}_{0} t + \frac{1}{2} g {t}^{2}$, where

$h$ - the height of the drop;
${v}_{0}$ - the initial speed of the balls, in your case ${v}_{0} = 0$;
$t$ - the time needed to cover the distance $h$;
$g$ - the gravitational acceleration - I'll use $g = {\text{10 m/s}}^{2}$;

Use the time it took the first ball to reach the ground to determine the value of $h$.

${h}_{1} = \frac{1}{2} \cdot 10 \text{m"/(cancel("s"^2)) * 4^2cancel("s"^2) = "80 m}$

Since the second ball is dropped 1 second after the first one, it's only 3 seconds into its free fall.

${t}_{2} = {t}_{1} - 1 = 4 - 1 = \text{3 s}$

The height it reached after 3 seconds will be

${h}_{2} = \frac{1}{2} \cdot 10 \text{m"/(cancel("s"^2)) * 3^2cancel("s"^2) = "45 m}$

This means that the distance between the first two balls will be

Delta_"1, 2" = 80-45 = color(green)("35 m")

In other words, the first ball dropped 35 meters more than the second ball.

The third ball was released 1 second after the second, which is equivalent to 2 seconds after the first ball.

${t}_{3} = {t}_{2} - 1 = 3 - 1 = \text{2 s}$

The distance travelled by the third ball will be

${h}_{3} = \frac{1}{2} \cdot 10 \text{m"/(cancel("s"^2)) * 2^2cancel("s"^2) = "20 m}$

This means that the distance between the second and third ball is

Delta_"2, 3" = 45 - 20 = color(green)("25 m")#

The fourth ball was released 1 second after the third ball, so

${t}_{4} = {t}_{3} - 1 = 2 - 1 = \text{1 s}$

The distance it covered is

${h}_{4} = \frac{1}{2} \cdot 10 \text{m"/(cancel("s"^2)) * 1^2cancel("s"^2) = "5 m}$

Thus, the distance between the third and the fourth ball is

${\Delta}_{\text{3, 4") = 20 - 5 = color(green)("15 m}}$