From the top of a tower a ball is thrown vertically upwards with an initial velocity u

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

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A ball is 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. Calculate1. The height of the tower2. The velocity of ball on reaching the ground. Take g =9.8 m / s