At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 20 m/s. When she lands, where will she find the ball? Ignore air resistance.

Use this kinematics equation to the describe the vertical motion of the ball as a function of time:

H (t) = X + Vt + 0.5At²

H (t) is the height of the ball

X is the initial height of the ball

V is the initial vertical velocity of the ball

A is the vertical acceleration of the ball

Given values:

X = 100m

V = - 2m/s

A = - 9.81m/s² (constant acceleration due to gravity near the surface of the earth)

Plug in the values:

H (t) = 100 - 2t - 4.905t²

To find a time t for when the ball hits the ground, plug in H (t) = 0 and solve for t:

100 - 2t - 4.905t²

t = 4.316s, - 4.724s

Reject the negative value because this doesn't make sense in the context of the question.

t = 4.316s

For the ball's horizontal motion under constant velocity:

X = VΔt

X is the horizontal distance traveled by the ball

V is the ball's velocity

Δt is the elapsed time

Given values:

V = 2.0m/s

Δt = 4.316s (from previous work)

Plug in and solve for X:

X = 2.0*4.316

X = 8.632m

The girl will find the ball 8.632m from where her hot air balloon will land.