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.

(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 𝑣2 − 𝑢2 = 2𝑔𝑠

We have, 02 − 492 = 2(−9.8)ℎ ⇒ ℎ = {49 \times 49\over2 \times9.8}= 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 𝑣 = 𝑢 + 𝑔𝑡

We get,

0 = 49 + (−9.8)𝑡 ⇒ 9.8𝑡 = 49 ⇒ 𝑡 = {49\over9.8} = 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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