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!
Hoping someone will help. Thank you!! Will try to solve the problem too. God bless fellow physics lover!
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I got 4.7059 lol idk if it is right pls help
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
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