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 effects of air are 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 correct answer is A because the time it takes for the rock to reach the maximum height of its trajectory is equal to the time it takes for the vertical component of its velocity to become zero. The vertical component of the initial velocity is $3.13 \sin 30.0° = 1.565 \mathrm {~m} / \mathrm { s }$. The acceleration due to gravity is $g = 9.80 \mathrm {~m} / \mathrm { s }^2$. Therefore, the time it takes for the rock to reach the maximum height of its trajectory is:$$t = \frac{v_f - v_i}{g} = \frac{0 - 1.565 \mathrm {~m} / \mathrm { s }}{-9.80 \mathrm {~m} / \mathrm { s }^2} = 0.160 \mathrm {~s}$$

A Boy Throws a Rock with an Initial Velocity Of 3.13