Fewpal
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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. 

1 Answer

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Best answer

(i) According to the equation of motion under gravity v2 − u2 = 2gs 

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 m/s and u = 49 m/s 

During upward motion, g = − 9.8 m s−2 

Let h be the maximum height attained by the ball. 

Hence, using 2 2 = 2 

We have, 02 − 492 = 2(−9.8)ℎ ⇒ ℎ = 49×49 / 2×9.8 = 122.5 m

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

We get, 

0 = 49 + (−9.8) ⇒ 9.8 = 49 ⇒ = 49 / 9.8 = 5 s 

But, 

Time of ascent = Time of descent 

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

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