A ball is released from a height and it reaches the ground in 3 s. If g = 9.8 m s-2, find — (a) the height from which the ball was released, (b) the velocity with which the ball will strike the ground.

Time (t) = 3 s

g = 9.8 m s^-2

initial velocity (u) = 0

(a) S = ut + 1/2 gt²

= 0 × 3 + 1/2 × 9.8 × 3²

= 0 + 44.1 m

= 44.1 m

Hence, the height from which the ball was released = 44.1 m

(b) From the equation of motion,

v² = u² – 2gs

where, v = final velocity

v² = 0² – 2 × 9.8 × 44.1

⇒ v² = 864.36

⇒ v = √864.36 = 29.4 m s^-1

Hence, the velocity with which the ball strikes the ground = 29.4 m s^-1