Question

A rock falls from a cliff with an acceleration of 13 m/s2 towards the ground. If the rock starts from rest, and it takes 3 s to hit the bottom, how high was the cliff in meters?

Answer 1

The cliff from which the rock was dropped is 58.5 meters high. This was calculated using the kinematic equation for uniformly accelerated motion, taking into account the acceleration of 13 m/s² and the fall time of 3 seconds.

Step-by-step explanation:

The question is asking to calculate the height of a cliff from which a rock is dropped and accelerates towards the ground at 13 m/s2, starting from rest and taking 3 seconds to hit the ground. To find the height, we can use the kinematic equation for uniformly accelerated motion.

Using the formula s = ut + ½at2, where s is the distance traveled, u is the initial velocity (which is 0 m/s since the rock starts from rest), a is the acceleration, and t is the time, we can plug in the values:

s = 0·t + ½·13·(32)

s = ½·13·9

s = 6.5·9

s = 58.5 meters

Therefore, the cliff is 58.5 meters high.