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 velocity of the sandbag at 0.250 s after its release. 7.45

Answer 1

The velocity of the sandbag after 0.250 seconds is 2.5475 m/s in upward direction relative to ground.

Step-by-step explanation:

Since the sand bag is released from a moving air balloon initially it will have the same velocity as the balloon itself.

Thus the case is that of motion under gravity.

The velocity after 0.250 seconds can be found by using first equation of kinematics as follows

`v=u+at`

we have u = 5.00 m/s

a=g=
`-9.81m/s^(2)` Since it acts opposite to the initial velocity.

Applying values in the equation we get

`v=5.00-9.81* 0.250=2.5475m/s`