Question

At a particular instant, a hot air balloon is 270 m in the air and descending at a constant speed of 3.5 m/s. At this exact instant, a girl throws a ball horizontally, relative to herself, with an initial speed of 25 m/s. When she lands, where will she find the ball? Ignore air resistance. (Find the distance, in meters, from the girl to the ball.)

Answer 1

Refer to the diagram shown below.

Neglect wind resistance.
g = 9.8 m/s².
Quantities are measured as positive upward.

When the ball is launched, it has a vertical velocity of - 3.5 m/s and a horizontal velocity of 25 m/s.

The time, t, for the ball to reach the ground is
(-3.5 m/s)*( t s) - 0.5*(9.8 m/s²)*( t s)² = - (270 m)
-3.5t - 4.9t² = - 270
4.9t² + 3.5t - 270 = 0
t² + 0.71433t - 55.102 = 0
Solve with the quadratic formula.
t = 0.5[-0.71433 +/- √220.918] = 7.0745 s or -7.788 s.
Reject negative time.

The horizontal distance traveled is
d = (25 m/s)*(7.0745 s) = 176.8625 m

Answer:
The horizontal distance from the girl to the ball will be 176.9 m (nearest tenth)