Question #226553

A ball is released from a window that is 10 meters above the ground. When the ball leaves your hand, it is moving 8m/s at an angle 300 below the horizontal. Ignoring the effect of air resistance, how far horizontally from the window will the ball hit the ground?


1
Expert's answer
2021-08-24T02:54:37-0400

Ux=ucosα=8cos30=6.92820m/sUy=usinα=8sin30=4m/sY=usinαt+12gt210=4t+6.92820t2Solving for t=0.95secHorizontal distance=Uxt=6.928200.95=6.58179mUx=ucosα=8cos30=6.92820m/sUy=usinα=8sin30=4m/sY=usinαt+21gt210=4t+6.92820t2Solvingfort=0.95secHorizontaldistance=Uxt=6.928200.95=6.58179mU_x=ucosα=8*cos30=6.92820m/s\\ U_y=usinα=8*sin30 =4m/s\\ Y=usinα*t+\frac{1}{2}gt^2\\ 10=4t+6.92820t^2\\ Solving \space for \space t=0.95sec\\ Horizontal \space distance=U_x*t=6.92820*0.95=6.58179mU x ​ =ucosα=8∗cos30=6.92820m/s U y ​ =usinα=8∗sin30=4m/s Y=usinα∗t+ 2 1 ​ gt 2 10=4t+6.92820t 2 Solving for t=0.95sec Horizontal distance=U x ​ ∗t=6.92820∗0.95=6.58179m


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