ch05-p056 - 56. To solve the problem, we note that the...

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which in turn gives 2 net, / (344.4 N)/(7.260 kg) 47.44 m/s . xx aF m == = Using Eq. 2-16 for constant-acceleration motion, the speed of the shot at the end of the acceleration phase is 22 2 0 2 (2.500 m/s) 2(47.44 m/s )(1.650 m) 12.76 m/s. x vv a x =+ Δ = + = (b) If 42 , θ then 2 net, 2 sin 380.0 N (7.260 kg)(9.80 m/s )sin 42.00 45.78 m/s , 7.260 kg x x F Fm g a mm −− ° = = and the final (launch) speed is 2 0 2 (2.500 m/s) 2(45.78 m/s )(1.650 m) 12.54 m/s. x a x Δ = + = (c) The decrease in launch speed when changing the angle from 30.00 ° to 42.00 ° is
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