at a particular instant a hot air balloon is 100m in the air and decending at a constant speed of 2m/s at this exact instant a girl throws a ball horizontally relative to herself with an initial speed of 20m/s when she lands where will she find the ball?

She will find the ball at a horizontal distance of 86.4 m from landed location

Explanation:

Consider the vertical motion of ball

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 2 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 100 m

Substituting

s = ut + 0.5 at²

100 = 2 x t + 0.5 x 9.81 xt²

4.905t² + 2t - 100 = 0

t = 4.32 s or t = - 4.72 s

After 4.32 seconds the ball reaches ground.

Now we need to find horizontal distance traveled by ball in 4.32 seconds.

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 20 m/s

Acceleration, a = 0 m/s²

Time, t = 4.32 s

Substituting

s = ut + 0.5 at²

s = 20 x 4.32 + 0.5 x 0 x 4.32²

s = 86.4 m

She will find the ball at a horizontal distance of 86.4 m from landed location