# College Physics 2.47 – Rock thrown from a cliff

## Solution:

### Part A

Refer to the figure below.

The known values are: $\text{t}=2.35\:\text{s};\:\text{y}=0\:\text{m};\:\text{v}_0=+8.00\:\text{m/s};\:\text{a}=-9.8\:\text{m/s}^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have

$0\:=\text{y}_0+\left(8.00\:\text{m/s}\right)\left(2.35\:\text{s}\right)+\frac{1}{2}\left(-9.80\:\text{m/s}^2\right)\left(2.35\:\text{s}\right)^2$

$y=8.26\:m$

Therefore, the cliff is 8.26 meters high.

### Part B

Refer to the figure below

The knowns now are: $y=0\:m;\:y_0=8.26\:m;\:v_0=-8.00\:m/s;\:a=-9.80\:m/s^2$

Based from the given values, we can use the equation $y=y_0+v_0t+\frac{1}{2}at^2$. Substituting the values, we have

$0\:m=8.26\:m+\left(-8.00\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2$

$4.9\text{t}^2+8\text{t}-8.26=0$

Using the quadratic formula to solve for the value of t, we have

$\displaystyle \text{t}=\frac{-8\pm \sqrt{\left(8\right)^2-4\left(4.9\right)\left(-8.26\right)}}{2\left(4.9\right)}$

$\text{t}=0.717\:\text{s}$