A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m s-2 find — (a) the initial velocity of the ball (b) the final velocity of the ball on reaching the ground and (c) the total time of journey of the ball.

Distance (s) = 20 m

g = 10 m s^-2

Initial velocity (v) = 0

v² = u² – 2gs (a = – g as movement is against gravity )

⇒ 0 = u² – 2 × 10 × 20

⇒ u² = 400

⇒ u = √400 = 20 m s^-1

Hence, the initial velocity of the ball = 20 m s^-1

(b) As we know, from the equation of motion,

When the ball starts falling after reaching the maximum height, it’s velocity (u) = 0

s = 20 m

g = 10 m s^-2

We know, v² = u² + 2gs

Substituting the values in the formula, we get,

v² = 0² + (2 × 10 × 20)

⇒ v² = 400

⇒ v = 20 ms^-1

Hence, final velocity of the ball on reaching the ground = 20 m s^-1

(c) As we know,

Initial velocity (u) = 20 m s^-1

gravity (g) = 10 m s^-2

Total time of journey (t) = 2u/g

⇒ t = 2 × 20/10 = 4 s

Hence, total time for which the ball stays in air = 4 s