Answer to Question #226553 in Civil and Environmental Engineering for camille

Question

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?

Answer 1

$U_x=ucosα=8*cos30=6.92820m/s\\ U_y=usinα=8*sin30 =4m/s\\ Y=usinα*t+\frac{1}{2}gt^2\\ 10=4t+6.92820t^2\\ Solving \space for \space t=0.95sec\\ Horizontal \space distance=U_x*t=6.92820*0.95=6.58179mU x ​ =ucosα=8∗cos30=6.92820m/s U y ​ =usinα=8∗sin30=4m/s Y=usinα∗t+ 2 1 ​ gt 2 10=4t+6.92820t 2 Solving for t=0.95sec Horizontal distance=U x ​ ∗t=6.92820∗0.95=6.58179m$