Question

A ball thrown vertically upwards from the top of a tower with an initial velocity of
19.6 m/s. The ball reaches the ground after 5 s. Calculate,
(i) the height of the tower.
(ii) the velocity of ball on reaching the ground. [g: 9.8 m/s²]

Answer 1

Let the time taken to reach the launch point again be t.

Initial velocity = 19.6 m/s

Final velocity when it again reaches the launch point going down = -19.6 m/s

Acceleration due to gravity = - 9.8 m/s^2

$v_f = v_i + at$

$-19.6 = 19.6 - 9.8 \times t$

$t = \frac{19.6 + 19.6 }{9.8}$

t = 4 s

Now, the time taken to reach the ground from the launch point = 5 - 4 = 1 s

Height of tower = H

Initial velocity = -19.6 m/s

$H = v_it + \frac{1}{2} at^2$

$H = -19.6 \times 1 - \frac{1}{2} \times 9.8 \times 1^2$

H = - 24.5 m

a) Hence, the height of the tower = 24 .5 m

b) Let the velocity at the bottom be v.

$v = v_i +at$

$v = - 19.6 - 9.8 \times 1$

v = - 29.4 m/s (negative because downward direction is taken as negative)

Hence, the velocity  at the bottom is - 29.4 m/s