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tpwatkin tpwatkin
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10 years ago
This is for a physics class... We are shooting a tennis ball from a catapult with an initial velocity of 8.8 meters per second, at an angle of 45 degrees, and the ball goes a distance of 6.9 meters. What's the air resistance on the ball?
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wrote...
10 years ago
I'm not entirely sure what parameter you are looking for in an answer.  If you want the air resistance as a force on the ball, it could take some careful calculation.  Usually a straight-forward answer is best, and so you could compare the distance that the ball should travel, vs. the observed distance.

The initial x-component of velocity is:

v0x = v*Cos(?) = 8.8*Cos(45) = 6.22 m/s

the initial y-component of velocity is:

v0y - v*Sin(?) = 8.8*Sin(45) = 6.22 m/s

Assuming the catapult releases the ball level with the floor, the time for the ball to reach zero velocity in the vertical direction (maximum height) would be:

vy(t) = v0y - gt = 0

t = v0y/g = 6.22 / 9.8 = .635 s

The time to impact with the floor (or flight time, tf) would be twice this,

tf = 2*.635 = 1.27 s

Calculate the distance that the ball should travel in the x direction:

x = v0x*t = 6.22 * 1.27 = 7.9 m

The air resistance reduces the ball horizontal travel by a distance of one meter.  If the point of release from the catapult is some distance above the floor, the expected travel distance would be greater, and the estimate of distance reduction would be a bit larger.
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