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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