Question

A ball is dropped from a height of 39m. The wind is blowing horizontally and imparts a constant acceleration of 1.2 m/s^2 on the ball. With what speed does the ball hit the ground?

Answer 1

here,
a = 1.2 m/s^2
H = 39 m

from third eqn of motion we have

V^2 - U^2 = 2 * a * H

V = sqrt(2 * a * H)
V = sqrt(2*1.2*39)
V = 9.6 m/s

time taken by ball wil be Equal to:
From first Eqn of motin we get,

T = V / a = 9.6 / 1.2
T = 8s

as wind is blowing perpendicular to ball falling direction therefore

D = 0.5 * a * t^2
D = 0.5* 1.2 *64
D = 38.4 m

So, velocity when ball hit ground will be

Vf = Sqrt(a * D)
Vf = sqrt(1.2 * 38.4 )

Vf = 6.7 m/s

The ball will hit the ground with velocity of 6.7 m/s