# Falling Objects| College Physics| Openstax| Problem 2.42|

#### Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

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

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

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

$y_5=-65.6\:m$

$v_5=v_0+at$

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

$v_5=-38.5:m/s$

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