Question

. A hot-air balloon is drifting straight downward with a constant speed of 2.40 m/s. When the balloon is 7.63 m above the ground, the balloonist decides to drop one of the ballast sandbags to the ground below. How much time elapses before the sandbag hits the ground?

Answer 1

Answer:

1.03 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

0 = 7.63 + (-2.40) t + ½ (-9.8) t²

0 = 7.63 - 2.40 t - 4.9 t²

Solve with quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

t = [ 2.40 ± √(2.40² - 4(-4.9)(7.63)) ] / -9.8

t = -1.52, 1.03

Since t can't be negative here, the sandbag hits the ground after 1.03 seconds.