Question

A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to be 3.2 seconds how deep is the canyon

Answer 1

Answer:

50.2 m

Explanation:

We can solve the problem by using the following SUVAT equation for the vertical position of the rock:

$y(t)=h+ut+\frac{1}{2}gt^2$

where

h is the initial height (the depth of the canyon), taking the bottom of the canyon as reference position

u = 0 is the initial velocity of the rock

t is the time

$g=-9.8 m/s^2$ is the acceleration of gravity

When the rock reaches the bottom, t = 3.2 s and y = 0. Substituting these numbers and solving for h, we find the depth of the canyon:

$h=\frac{1}{2}gt^2 = -\frac{1}{2}(-9.8)(3.2)^2=50.2 m$