Question

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

Answer 1

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 downwards direction as positive (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 negative sign indicates that it is pointing upwards.