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.)

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

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

The best answer i can think of is b

So since the snail is moving at 3.6  miles per hour you multiply by 5280 to change it into feet per hour. so you get 36*528 ft per hour. since there are 3600 seconds in an hour, you divide 36*528 by 3600. so simplifying we get 528/100 which is 5.28 ft per second