# Question 62d9c

Sep 30, 2015

$\text{58.8 m}$

#### Explanation:

You can break the motion of the ball into two parts

• moving upwadrs from the height of the tower while being decelerated by the gravitational acceleration
• free falling from maximum height.

At maximum height, the velocity of the ball is equal to zero. This means that you an say

underbrace(v_"top"^2)_(color(blue)(=0)) = v_0^2 - 2 * g * h" ", where

$h$ - the height travelled upward from the initial position.

You get that

h = v_0^2/(2 * g) = (19.6""^2"m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^2))))/(2 * 9.8color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = "19.6 m"

The time it took to reach maximum height, i. e. travel $h$ upward, is

underbrace(v_"top")_(color(blue)(=0)) = v_0 - g * t_"up"

${t}_{\text{up" = v_0/g = (19.6color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(9.8color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "2 s}}$

So, it climbs for two seconds, and its total flight time is six seconds, then that must mean that it fell from maximum height in four seconds

${t}_{\text{down" = t_"total" - t_"up}}$

${t}_{\text{down" = "6 s" - "2 s" = "4 s}}$

This means that maximum height, $H$, is

H = underbrace(v_"top")_(color(blue)(=0)) * t_"down" + 1/2 * g * t_"down"^2

$H = \frac{1}{2} \cdot 9.8 \text{m"/color(red)(cancel(color(black)("s"^2))) * 4""^2color(red)(cancel(color(black)("s"^2))) = "78.4 m}$

The height of the tower was

${h}_{\text{tower}} = H - h$

h_"tower" = "78.4 m" - "19.6 m" = color(green)("58.8 m")#