A hot-air balloonist, rising vertically with a constant velocity of magnitude v=

Question

A hot-air balloonist, rising vertically with a constant velocity of magnitude v= 5.00 m/s, releases a sandbag at an instant when the balloon is a height h= 40.0 m above the ground . After it is released, the sandbag is in free fall. For the questions that follow, take the origin of the coordinate system used for measuring displacements to be at the ground, and upward displacements to be positive. Take the free fall acceleration to be g= 9.80 m/s^2.

1. How many seconds after its release will the bag strike the ground?

Explanation / Answer

In a previous part to this question, I already solved for you the fact that the bag will rise 1.276 m before it begins to fall.

The total distance of fall will be 41.276 m

To find the final velocity when it hits, use the formula

vf2 = vo2 + 2ad

vf2 = (0) = (2)(9.8)(41.276)

vf = 28.44 m/s

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