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

Question

creativ13

18 days 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:

kicyunya [2.2K] · Answer 1 · 2026-03-08T05:56:11+03:00

Conclusion:

B) 225 m/s^2

Details:

The rock impacts the ground at 15m/s and penetrates 50cm=0.5m into the ground until it comes to a halt.

Assuming the acceleration is constant, the formula we will use is $v^2=v_0^2+2ad$ , which for acceleration is expressed as:

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

Assuming the downward direction is taken as positive (the direction of travel), both the initial velocity and displacement will be considered positive, and we substitute our values aligned with this movement:

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

The negative sign denotes upward direction.