QuestionMay 31, 2025

Jill is at the edge of a 30.0 m high cliff and throws a rock with a velocity of 10.0m/s at an angle of 20.0 degrees. How long will it take the rock to hit the water? 285 s 3.23 s 3.56 s 3.86 s

Jill is at the edge of a 30.0 m high cliff and throws a rock with a velocity of 10.0m/s at an angle of 20.0 degrees. How long will it take the rock to hit the water? 285 s 3.23 s 3.56 s 3.86 s
Jill is at the edge of a 30.0 m high cliff and throws a rock with a velocity of 10.0m/s at an angle of 20.0 degrees. How long will it take the rock to hit the water?
285 s
3.23 s
3.56 s
3.86 s

Solution
3.7(248 votes)

Answer

3.23 s Explanation 1. Calculate initial velocity components v_{x} = v \cdot \cos(\theta) = 10.0 \cdot \cos(20^\circ); v_{y} = v \cdot \sin(\theta) = 10.0 \cdot \sin(20^\circ). 2. Determine time to hit the water Use y = v_{y} \cdot t + \frac{1}{2} \cdot g \cdot t^2. Set y = -30.0 m (downward), solve for t: -30.0 = v_{y} \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2.

Explanation

1. Calculate initial velocity components<br /> $v_{x} = v \cdot \cos(\theta) = 10.0 \cdot \cos(20^\circ)$; $v_{y} = v \cdot \sin(\theta) = 10.0 \cdot \sin(20^\circ)$.<br />2. Determine time to hit the water<br /> Use $y = v_{y} \cdot t + \frac{1}{2} \cdot g \cdot t^2$. Set $y = -30.0$ m (downward), solve for $t$: $-30.0 = v_{y} \cdot t - \frac{1}{2} \cdot 9.8 \cdot t^2$.
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