104_Physics ProblemsTechnical Physics

104_Physics ProblemsTechnical Physics - Chapter 4 P4.56 105...

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Chapter 4 105 P4.56 Using the range equation (Equation 4.14) R v g ii = 2 2 sin( ) θ the maximum range occurs when i 45 , and has a value R v g i = 2 . Given R , this yields vg R i = . If the boy uses the same speed to throw the ball vertically upward, then R g t y =− and yg R t gt 2 2 at any time, t . At the maximum height, v y = 0, giving t R g = , and so the maximum height reached is R R g g R g R RR max F H G I K J =− = 22 2 2 . P4.57 Choose upward as the positive y -direction and leftward as the positive x -direction. The vertical height of the stone when released from A or B is y i =+ ° = 1 50 1 20 30 0 2 10 .. . m . m sin af (a) The equations of motion after release at A are vv g t t yt t xt yi xi A = = = sin cos 60 0 1 30 9 80 60 0 0 750 210 130 490 0750 2 . m s m s .+ . . m . m ej a f v i BA v i 30 ° 30 ° 1.20 m FIG. P4.57 When y = 0, t = −± + = 1 30 1 30
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