Question

A hot-air balloonist, rising vertically with a constant speed of 5.00 m/s releases a sandbag at the instant the balloon is 40.0 m above the ground. After it is released, the sandbag encounters no appreciable air drag. Compute the position of the sandbag at 0.250 s after its release.

Answer 1

Answer:

Y = 40.94m

Explanation:

The initial speed of the sandbag is the same as the balloon and so is its position, so:

$Y = Yo + Vo*t-\frac{g*t^2}{2}$

Replacing these values:

Yo = 40m     Vo = 5m/s     g = 9.81m/s^2     t = 0.25s

We get the position of the sandbag:

$Y = 40+5*(0.25)-\frac{9.81*(0.25)^2}{2}$

Y = 40.94m