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pathosdebater
A baseball player bats a ball at a speed of 90.0 ft/sec at an angle of 40.0 degrees in an indoor baseball stadium. The roof is 90.0 feet high. What is the total time that the baseball is in the air?
Because of the symmetry of the problem, the total time of flight will be twice the time it takes for the ball to reach it's highest point. To find this, we need the vertical component of the speed. We get this as shown in the drawing:|dw:1353078314614:dw|So \[v_y=90\sin 40=57.85\]Now we have a one dimensional problem of a ball going up with a speed of 57.85m/s under a constant force of gravity F=-mg Can you take it from here?\[v_f=v_i+at\]Here vf=0, vi=57.85 and a=-9.81. Solve for t. Don't forget to multiply this by two to get the total time of flight.
Watch out for that roof
if you stay in feet, the acceleration due to gravity is 32 ft/sec^2
@phi didn't see that, thanks. I thought everything was in metric...ooops. Yeah, use 32 ft/s...and watch out for the roof
I calculated the maximum height to be around 50 something ft... Is it higher? Cause you guys are talking about watching out for the roof...
I think the roof is high enough.
If you thought the 90 ft/sec were straight up, you could hit the roof, but that is not what is going on here.
I do have to calculate the height when y=0 (which is at the pinnacle of the parabola)... Basically just the highest point the ball ever goes
What equation should I use to find the max height?
eseidl 's write up shows how to find t, the time it takes the ball to slow down to 0 (in the vertical direction) That is the fastest way to the answer. just double that t to get the total time in the air
If you want distance (you don't need it here) \[ s = v_0 t + \frac{1}{2} g t^2 \] where g is -32 ft/sec^2 and v0 is 90 sin(40)
@pathosdebater yeah 50 something is what I got for max height as well so the roof won't interfere with the balls trajectory