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)

What formula  this one  (x_/

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