Falling Objects| College Physics| Openstax| Problem 2.48|

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long does he have to get out of the way if the shot was released at a height of 2.20 m, and he is 1.80 m tall?

SOLUTION:

The known values are

y_0=2.20\:m;\:y=1.80\:m;\:v_0=11.0\:m/s;\:and\:a=-9.80\:m/s^2

The applicable formula is \Delta y=v_0t+\frac{1}{2}at^2.

1.80\:m-2.20\:m=\left(11.0\:m/s\right)t+\frac{1}{2}\left(-9.80\:m/s^2\right)t^2

-0.40\:m=\left(11.0\:m/s\right)t-\left(4.90\:m/s^2\right)t^2

4.90t^2-11t-0.40=0

Using the quadratic formula solve for t, we have

\:t=\frac{-\left(-11\right)\pm \sqrt{\left(-11\right)^2-4\left(4.90\right)\left(-0.40\right)}}{2\left(4.90\right)}

t=2.28\:sec\:\:\:\:\:or\:\:\:\:\:\:t=-0.04

Therefore, t=2.28\:s

 

 

 

 

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