Question

A small ball is thrown vertically upwards from the top of a tower with an initial velocity of 20 m/s. If the ball took a total of 6 s to reach the ground level, determine the height of the tower.

Answer 1

Split the time into several parts. First, calculate how long it will take to reach the highest point of trajectory if the body is launched vertically at 20 m/s:

$$t_1=\frac{v}{g}.$$

Then the body falls from the maximum height $H$ and this takes time $t_2=t-t_1$:

$$H=\frac{gt_2^2}{2}=\frac{g(t-t_1)^2}{2}=\frac{g(t-v/g)^2}{2}.$$

The height above the tower that the body reached is

$$y=\frac{v^2}{2g}.$$

Thus, the height of the tower is

$$h=H-y=\frac{g(t-v/g)^2}{2}-\frac{v^2}{2g},\\\space\\ h=\frac{9.8(6-20/9.8)^2}{2}-\frac{20^2}{2\cdot9.8}=56.4\text{ m}.$$