Falling Objects| College Physics| Openstax| Problem 2.41|

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y_o=0.

SOLUTION:

The given known quantities are 

a=-9.8\:m/s^2

y_o=0\:m

v_o=+15\: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(15.0\: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=6.28\:m

v_1=v_0+at_1

v_1=\left(15.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(0.500\:s\right)

v_1=10.1\:m/s

Part b

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

y_2=0\:m+\left(15.0\: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=10.1\:m

v_2=v_0+at_1

v_2=\left(15.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(1.00\:s\right)

v_2=5.20\:m/s

Part c

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

y_3=0\:m+\left(15.0\: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=11.5\:m

v_3=v_0+at_1

v_3=\left(15.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(1.50\:s\right)

v_3=0.300\:m/s

Part d

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

y_4=0\:m+\left(15.0\: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=10.4\:m

v_4=v_0+at_1

v_4=\left(15.0\:m/s\right)+\left(-9.8\:m/s^2\right)\left(2.00\:s\right)

v_4=-4.60\:m/s

 

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