A track star in the broad jump goes into the jump at 12.0 m/s and launches himself at 26.0° above the horizontal.How long is he in the air before returning to Earth? (g = 9.8 m/s²) A)0.5 s B)1.1 s C)2.1 s D)2.2 s E)1.2 s

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Quizplus · Accepted Answer

The time of flight in projectile motion is given by $T = \frac{2u\sin\theta}{g}$, where $u$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity. Substituting $u = 12.0 \, \text{m/s}$, $\theta = 26.0^\circ$, and $g = 9.8 \, \text{m/s}^2$, we get $T = \frac{2 \times 12.0 \times \sin(26.0^\circ)}{9.8} \approx 1.1 \, \text{s}$.

A Track Star in the Broad Jump Goes into the Jump