A ball is thrown vertically upwards with an initial velocity of 49 m s-1. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g = 9.8 m s-2)

Initial velocity (u) = 49 m s^-1

Final velocity (v) = 0

g = 9.8 m s^-2

From the equation of motion,

v² = u² – 2gh

⇒ 0² = 49² – 2 × 9.8 × h

⇒ 19.6 h = 49²

⇒ h = 2401/19.6 = 122.5 m

Hence, maximum height attained = 122.5 m

(ii) As we know, the total time of the journey is,

t = 2u/g = 2 × 49/9.8 = 10 s

Hence, the time taken by the ball before it reaches the ground again = 10 s