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Lunna [17]
3 years ago
14

A rock hits the ground at a speed of 15 m/s and leaves a hole 50 cm deep. After it hits the ground, what is the magnitude of the

rock's (Assumed)uniform acceleration? A) 112.5 m/s^2
B) 225 m/s^2
C) 225
D) 127.5 m/s^2
Physics
1 answer:
garik1379 [7]3 years ago
4 0

Answer:

B) 225 m/s^2

Explanation:

The rock hits the ground at 15m/s and travels 50cm=0.5m through the ground until it stops.

The acceleration is supposed to be uniform, so the formula we have to use is v^2=v_0^2+2ad, which for acceleration is:

a=\frac{v^2-v_0^2}{2d}

Taking the <em>downwards direction as positive</em> (the direction of traveling, so the initial velocity and displacement will be positive), substituting our values for that movement we have:

a=\frac{v^2-v_0^2}{2d}=\frac{(0m/s)^2-(15m/s)^2}{2(0.5m)}=-255m/s^2

Where the <em>negative sign indicates that it is pointing upwards.</em>

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