Question

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description: show all you work, including units, on separate paper. follow the \problem solving method\. this final exam is cumulative and covers material from the entire course.
instructions: from the list of choices, select the one best answer.
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question 19 of 62
question 19
1.6 points save answer
a boy throws a rock with an initial velocity of 2.15 m/s at 30.0° above the horizontal. how long does it take for the rock to reach the maximum height of its trajectory?
○ 0.303 s
○ 0.110 s
○ 0.215 s
○ 0.194 s

Explanation:

Step1: Find vertical component of velocity

The initial velocity is \( v_0 = 2.15 \, \text{m/s} \) at an angle \( \theta = 30.0^\circ \). The vertical component \( v_{0y} \) is given by \( v_{0y}=v_0\sin\theta \). So, \( v_{0y}=2.15\sin(30.0^\circ) \). Since \( \sin(30.0^\circ) = 0.5 \), we have \( v_{0y}=2.15\times0.5 = 1.075 \, \text{m/s} \).

Step2: Use kinematic equation for time to max height

At maximum height, the vertical velocity \( v_y = 0 \). The kinematic equation is \( v_y = v_{0y}-gt \), where \( g = 9.8 \, \text{m/s}^2 \). Solving for \( t \): \( 0 = 1.075 - 9.8t \), so \( t=\frac{1.075}{9.8} \approx 0.110 \, \text{s} \).

Answer:

0.110 s (corresponding to the option "0.110 s")