A varsity player is attempting to make a shot. The ball leaves the hands of the player at an angle of 50 degrees to the horizontal at an elevation of 2 meters above the floor. The skillful player makes the shot with the ball travelling precisely through the center of the ring (8 meters from the player and 3 meters above the floor). To loud cheers, calculate the speed at which the ball left the hands of the player. Sketch the problem!
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Hoping someone will help. Thank you!! Will try to solve the problem too. God bless fellow physics lover!
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Vox = Vocostheta -> 8=Vocos(50degrees) -> Vo=12.446m/s
Then I plug it to X=Vot+1/2at^2 a is always 0 right? So it becomes X=Vot -> X=12.446(t) -> t=(0.643s)
Then I plug the time to the y equation of free fall.. Delta Y is 1 right? Because 3-2=1
Y=Vot-1/2gt -> 1=Vo(0.643)-1/2(9.8)(0.643)^2 Vo=4.7059m/s