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Ch 02: Motion Along a Straight Line
Chapter 2, Problem 2

A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases a sandbag at an instant when the balloon is 40.0 m above the ground (Fig. E2.44). After the sandbag is released, it is in free fall. (d) What is the greatest height above the ground that the sandbag reaches? A hot-air balloonist releases a sandbag from 40 m height, rising at 5 m/s.
Illustration of a hot-air balloonist releasing a sandbag at 40 m height.

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1
Identify the initial conditions: The sandbag is released from a height of 40.0 m above the ground with an initial upward velocity of 5.00 m/s.
Use the kinematic equation for vertical motion to find the maximum height: v^2 = u^2 + 2as, where v is the final velocity (0 m/s at the highest point), u is the initial velocity (5.00 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and s is the displacement.
Rearrange the equation to solve for the displacement (s): s = (v^2 - u^2) / (2a).
Calculate the displacement (s) from the point of release to the highest point.
Add the displacement (s) to the initial height (40.0 m) to find the greatest height above the ground that the sandbag reaches.

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