# 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 $y_5=y_0+v_ot+\frac{1}{2}at^2$ $y_5=0\:m+\left(-14.0\:m/s\right)\left(2.50\:s\right)+\frac{1}{2}\left(-9.8\:m/s^2\right)\left(2.50\:s\right)^2$ $y_5=-65.6\:m$ $v_5=v_0+at$ $v_4=\left(-14.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(2.50\:s\right)$ $v_5=-38.5:m/s$