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.

Answer 1

`v = 2.55\ m / s` ↑ upwards

Step-by-step explanation:

Take the direction up (where the hot-air balloonist moves) as the positive direction, and take the direction down (contrary to the direction of the hot-air balloonist) as the negative direction.

Then we propose the kinematic equation for the speed of the sandbag:

`v (t) = v_0 + at`.

Where

`v_0` is the initial velocity of the sandbag

a is the acceleration of the sandbag

t is the time in seconds.

In this case

a is the gravitational acceleration of
`9.8\ m / s ^ 2`

`v_0 = 5\ m / s`

`t = 0.250\ s`

Now we substitute these values into the formula and solve for
`v(t)`

`v = 5 + (-9.8) (0.250)\\\\v = 5 -9.8 (0.250)\\\\`

`v = 2.55\ m / s` ↑ upwards