Question

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. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9·8 m/s^2.​

Answer 1

Answer:

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Answer 2

Solution :-

As the ball is thrown vertically upwards from the top of a tower with an initial velocity of 19·6 m/s, the time taken by the ball to reach the maximum height is

t = u / g

t = 19.6/9.8 sec

t = 2 seconds

And the height it reached with respect to the roof

h = u² / (2g)

= 19.6² / (2 x 9.8)

= 19.6 m ………(1)

Now as per question the ball takes 5 secs to reach the ground after it was thrown upwards.

This means the time taken by the ball to come down from its max height position to the ground = 5 – 2 = 3 seconds.

Therefore, during downward movement the ball traversed a distance say

H = ½ g t²

= ½ x 9.8 x 3²

= 44.1 m

Now to get the height of the tower we have to deduct the height it rose with respect to the roof from this 44.1 m.

So,

Height of the tower = H – h

= 44.1 – 19.6

= 24.5 m

[ Answer of (i) ]

The velocity of the ball on reaching the ground

V = gt

= 9.8 x 3

= 29.4 m/s

[ Answer of (ii) ]