Question 1044694: A plastic ball that is released with a velocity of 15 m/s stays in the air for 2.0 sec.
a.) At what angle with respect to the horizontal was it released?
b.) What was the maximum height achieved by the ball?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Time of flight is 2 v sub i * sin theta/g, g is 9.8 m/sec^2, t=2 sec
19.6 m/sec=2(15 m/sec)* sin theta
19.6/30=sin theta and it will be dimensionless.
0.65333=sin theta, and theta is 40.79 or 40.8 degrees.
h=v sub i ^2*sin^2 theta/g
=225 m^2/sec^2*0.427/9.8 m/sec^2, units will be meters.
without rounding until the end, this is 9.799 or 9.8 m.
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