**(i) **According to the equation of motion under gravity v^{2} − u^{2} = 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 𝑣^{2} − 𝑢^{2} = 2𝑔𝑠

We have, 0^{2} − 49^{2} = 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