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

Question

(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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Salim Ahmed Kawsar · Answer 1 · Feb 05, 2023 15:29:48

(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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