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. Compute the position of the sandbag at a time 1.50 s after its release.

2. Compute the velocity of the sandbag at a time 1.50 s after its release.

Explanation / Answer

V = 5 - 9.8*1.5 v = - 9.7 m/s that is downwards s = 5*1.5- 0.5*9.8*1.5^2 S = -3.5250 so, the particle is 40-3.5250 = 36.4750 m

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