A boy throws a rock with an initial velocity of $3.13 \mathrm {~m} / \mathrm { s }$ at 30.0° above the horizontal. How long does it take for the rock to reach the maximum height of its trajectory if air resistance is negligibly small and g = 9.80 m/s²? A)0.160 s B)0.282 s C)0.313 s D)0.441 s

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Quizplus · Accepted Answer

The time to reach the maximum height is independent of the horizontal component of the initial velocity. The vertical component of the initial velocity is:$$v_{y,i} = v_i \sin 	heta = (3.13 \mathrm{~m/s}) \sin 30.0° = 1.565 \mathrm{~m/s}$$The time to reach the maximum height is:$$t = \frac{v_{y,i}}{g} = \frac{1.565 \mathrm{~m/s}}{9.80 \mathrm{~m/s^2}} = 0.160 \mathrm{~s}$$

A Boy Throws a Rock with an Initial Velocity Of 3.13