# College Physics 2.42 – A rock thrown straight down

## Solution:

The given known quantities are

$a=-9.8\:m/s^2$

$y_o=0\:m$

$v_o=-14\:m/s$

To compute for the displacement, we use the formula

$y=y_0+v_0t+\frac{1}{2}at^2$,

and to compute for the final velocity, we use the formula

$v=v_0+at$

### Part A

$y_1=y_0+v_ot+\frac{1}{2}at^2$

$y_1=0\:m+\left(-14.0\:m/s\right)\left(0.500\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(0.500\:s\right)^2$

$y_1=-8.23\:m$

$v_1=v_0+at$

$v_1=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(0.500\:s\right)$

$v_1=-18.9\:m/s$

### Part B

$y_2=y_0+v_ot+\frac{1}{2}at^2$

$y_2=0\:m+\left(-14.0\:m/s\right)\left(1.00\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(1.00\:s\right)^2$

$y_2=-18.9\:m$

$v_2=v_0+at$

$v_2=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(1.00\:s\right)$

$v_2=-23.8\:m/s$

### Part C

$y_3=y_0+v_ot+\frac{1}{2}at^2$

$y_3=0\:m+\left(-14.0\:m/s\right)\left(1.50\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(1.50\:s\right)^2$

$y_3=-32.0\:m$

$v_3=v_0+at$

$v_3=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(1.50\:s\right)$

$v_3=-28.7\:m/s$

### Part D

$y_4=y_0+v_ot+\frac{1}{2}at^2$

$y_4=0\:m+\left(-14.0\:m/s\right)\left(2.00\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(2.00\:s\right)^2$

$y_4=-47.6\:m$

$v_4=v_0+at$

$v_4=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(2.00\:s\right)$

$v_4=-33.6\:m/s$

### Part E

$y_5=y_0+v_ot+\frac{1}{2}at^2$

$y_5=0\:m+\left(-14.0\:m/s\right)\left(2.50\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(2.50\:s\right)^2$

$y_5=-65.6\:m$

$v_5=v_0+at$

$v_4=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(2.50\:s\right)$

$v_5=-38.5:m/s$