Ahot-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.

40.9 m

39.7 m

41.3 m

40.3 m

First option

The position of the sandbag at 0.250 s after its release is 40.9 m

Explanation:

We want to know the height of the balloon as a function of time, after 0.250s

Then we use the following equation of motion:

In this equation

is the speed

is gravitational acceleration

is the initial height

is the initial velocity.

If the sandbag is released 40 meters above the ground, then the initial height is 40 meters.

As the balloon rises at a constant speed, then the initial velocity of the balloon is 5 m/s

Now we substitute these values in the formula

Now we look for:

a=d/t

a="triangle"v/t

explanation:

a = acceleration