Physics
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 position of the sandbag at 0.250 s after its release.
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The position of the sandbag 0.250 s after its release would be 10 m above the ground. This is due to the fact that the sandbag has a constant speed given by the balloonist (5.00 m/s) and the time it takes the sandbag to reach this position is 0.250 s. This means that the sandbag has traveled 5 m (5.00 m/s * 0.250 s), and since it was initially released at 40 m above the ground, the sandbag would be positioned 10 m above the ground (40 - 5 = 35 m).
