Question

A ball is thrown vertically upwards from the top of a
tower with an initial velocity of 19.6 ms.per 2 The ball
reaches the ground after 5 s. Calculate : (i) the height
of the tower, (ii) the velocity of ball on reaching the
ground. Take g = 9.8 ms per 2
​

Answer 1

Let x be the height of the tower. Let h be the distance from the top of the tower to

the highest point

nitial velocity u = 19.6 m/s

g = 9.8 m/s2

At the highest point, velocity 0

Using the third equation of motion,

v2 - u? = 2gh

Or, - (19.6) 2 = 2 (-9.8) h

Or, h = 19.6 m

If the ball takes time t, to go to the highest point from the top of building, then for the upward journey from the relation, v = u gt,

0 = 19.6 (9.8) (t)

(ii) Let us consider the motion for the part (x+h) Time taken to travel from highest point to the

ground = (5 2) = 3s

Using the equation s = ut + (1/2) gt2

We get,

(x + h) = 0 + (1/2) (9.8) (3)2

Or, (x + 19.6) = 44.1 m

Or, x = 44.1 19.6 = 24.5 m

Thus, height of the tower 24.5 m

(iii) Let v be the velocity of the ball on reaching

the ground.

Using the relation, v u + gt

We get:

v = 0 + (9.8) (3)

Or, v = 29.4 m/s