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.

 

CLICK HERE FOR 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

You can now buy the complete solution manual of College Physics by Openstax. Just click on the button you can see at the right portion of this post.

 

Advertisements