ps3b7 - Ph 111.01 Fall, 2008 Solutions, Problem set 3b (7)...

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Ph 111.01 Fall, 2008 Solutions, Problem set 3b (7) Problem 1. A cannonball is fired out of a cannon at [50] m/s at [45] degrees above horizontal, on level ground. (a) How long will the cannonball be in the air? (b) What is the maximum height above the ground that it will reach? (c) How far will the cannonball travel horizontally? Solution. We have to make a drawing that shows the components of the initial velocity. () ( ( 00 cos 50 m/s cos 45 sin 50 m/s sin 45 x y vv ) ) 35.36 m/s 35.36 m/s = = θ = = 0 yy t (They happen to be the same because the angle is exactly 45 degrees.) x y v 0 h R θ v 0y v 0x (a) We will calculate the time to get to the top of the path, using , with v vv g =− y = 0 and v 0y = 35.36 m/s. This gives ( ) 0 2 35.36 m/s 3.61 s 9.8 m/s y v t g == = The total time in the air is twice this, or 7.22 s. (b) We will use the equation for y , with y 0 = 0 and v 0y = 35.36 m/s: ( ) 2 1 0 2 2 2 1 2 35.36 m/s 3.61 s 9.8 m/s 3.61 s y yv t g t 63.79 m = 0 x (c) The horizontal distance traveled is just the horizontal velocity, equal to v 0x , times the total time in the air. ( ) 35.36 m/s 7.22 s 255.3 m xv t = y θ 50 ° v 0 : 5 m/s x v 0y 3.83 m/s Problem 2. A ball is thrown off a [37] m high cliff, at [7] m/s, 50 degrees above the horizontal.
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ps3b7 - Ph 111.01 Fall, 2008 Solutions, Problem set 3b (7)...

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