A track star in the broad jump goes into the jump at 12 m/s and launches himself at 20 $\degree$above the horizontal.How long is he in the air before returning to Earth? (g = 9.8 m/s²) A)0.42 s B)0.84 s C)1.25 s D)1.68 s

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Quizplus · Accepted Answer

The time in the air is determined by the vertical component of the velocity. The vertical velocity is 12 m/s * sin(20°). Using the formula t = 2 * (v * sin(θ)) / g, we find t = 2 * (12 * sin(20°)) / 9.8 ≈ 0.84 s.

A Track Star in the Broad Jump Goes into the Jump