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

Y = 40.94m

Step-by-step explanation:

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

`Y = Yo + Vo*t-(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)-(9.81*(0.25)^2)/(2)`

Y = 40.94m