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A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises.

(ii)the total time it takes to return to the surface of the earth.

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(i) 122.5 m (ii) 10 s According to the equation of motion under gravity: v2 − u2 = 2 gs Where, u = Initial velocity of the ball v = Final velocity of the ball s = Height achieved by the ball g = Acceleration due to gravity At maximum height, final velocity of the ball is zero, i.e., v = 0 u = 49...
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(i) 122.5 m (ii) 10 s

According to the equation of motion under gravity:

v2 − u2 = 2 gs

Where,

u = Initial velocity of the ball

v = Final velocity of the ball

s = Height achieved by the ball

g = Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., v = 0

= 49 m/s

During upward motion, g = − 9.8 m s−2

Let h be the maximum height attained by the ball.

Hence,

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

But,

Time of ascent = Time of descent

Therefore, total time taken by the ball to return = 5 + 5 = 10 s

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