Question

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s^-1. The ball reaches the ground after 5 s. Calculate: (i) The height of the tower, (ii) The velocity of the ball on reaching the ground. Take g= 9.8 ms^-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 - u² = 2gh

Or, - (19.6) ² = 2 (-9.8) h

Or, h = 19.6 m

 

If the ball takes time t₁ 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) (t₁)

Or, t₁ = 2s

(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