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A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strike...Asked by Lindsay
A regulation volleyball court is L = 18.0 m long and a regulation volleyball net is d = 2.43 m high. A volleyball player strikes the ball a height h = 1.76 m directly above the back line, and the ball's initial velocity makes an angle q = 55° with respect to the ground. At what initial speed must the ball be hit so that it lands directly on the opponent's back line?
Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court?
What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???
Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court?
What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???
Answers
Answered by
drwls
<<Would my equation for the x-dir. be Vcos55*T = 18, since it's the full length of the court?
YES VT = 31.38 m
<<What would my height for the y-dir. be?>>
Vsin55*T - (g/2)T^2 = -1.76
sin55*31.38 - 4.9 T^2 = -1.76
25.71 + 1.76 = 4.9 T^2
T = 2.37 s
V = 13.2 m/s
After you get V by solving the two simultaneous equations, make sure that the net is also cleared by calculating the height at which it passes by the net.
YES VT = 31.38 m
<<What would my height for the y-dir. be?>>
Vsin55*T - (g/2)T^2 = -1.76
sin55*31.38 - 4.9 T^2 = -1.76
25.71 + 1.76 = 4.9 T^2
T = 2.37 s
V = 13.2 m/s
After you get V by solving the two simultaneous equations, make sure that the net is also cleared by calculating the height at which it passes by the net.
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