find the height of the racket ball when it left the racket??
Homework Answers
so 1.95=0+4.87*t + 0.5 * 0 * t^2
so t=1.95/4.87=0.4
now y direction y=y0 + v0y t + 1/2 a t^2
0=y0+0*.4 + 0.5* (-9.8) * .4^2
so y0=0.5*9.8*0.4^2=0.784 m
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