Question

a boy throws a rock with an initial velocity of 2.15 m/s at 30.0° above the horizontal. how far horizontally does it travel when it reaches it maximum height?
0 0.237 m
0 0.205 m
0 0.102 m
0 0.458 m

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 \( v_{0y} = v_0 \sin\theta \). So \( v_{0y} = 2.15 \sin(30.0^\circ) = 2.15 \times 0.5 = 1.075 \, \text{m/s} \).

Step2: Time to reach max height

At maximum height, vertical velocity \( v_y = 0 \). Using \( v_y = v_{0y} - gt \), solve for \( t \): \( 0 = 1.075 - 9.8t \), so \( t = \frac{1.075}{9.8} \approx 0.1097 \, \text{s} \).

Step3: Horizontal component of velocity

Horizontal component \( v_{0x} = v_0 \cos\theta = 2.15 \cos(30.0^\circ) \approx 2.15 \times 0.8660 \approx 1.862 \, \text{m/s} \).

Step4: Horizontal distance

Horizontal distance \( x = v_{0x} \times t \). Substitute values: \( x = 1.862 \times 0.1097 \approx 0.205 \, \text{m} \).

Answer:

0.205 m