A track star in the broad jump goes into the jump at 10 m/s and launches himself at 20 $\degree$ above the horizontal. How long is he in the air before returning to Earth? Note: g = 9.8 m/s². A) 0.35 s B) 0.42 s C) 0.70 s D) 0.89 s

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Quizplus · Accepted Answer

The time in the air is determined by the vertical component of the velocity. Using the formula $ t = \frac{2v_0 \sin(\theta)}{g} $, where $ v_0 = 10 \, \text{m/s} $ and $ \theta = 20^\circ $, the time is approximately 0.70 s.

A Track Star in the Broad Jump Goes into the Jump