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) Initial velocity (u) = 49 m/s

Final velocity (v) = 0 m/s

Height achieved by the ball (s) = ?

Acceleration due to gravity (g) = − 9.8 m s⁻²

Let h be the maximum height attained by the ball.

We know, v² − u² = 2gs

Hence, using v² − u² = 2gh

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