Suppose the height of the net is 0.800 m , and the court's border is 11.2 m from the bottom of the net. Use g =10 m/s2 and find the maximum speed of the horizontally moving ball clearing the net..

Answer 1

Ok, it is a projectile problem.

a. find the initial vertical speed. The at the mazimum height, its vertical speed is zero.
vf^2=vi^2+2ah in the vertical
0=vi^2)+2 (-10)h
vi=sqrt(20*.8) =sqrt10m/s=4m/s
now, that is the initial vertical speed.
ok, next is the time it takes the ball to reach the maximum point.
In the vertical=
vf=vi-gt
0=4m/s-10t or t=.4sec

so the horizontal velocity is now 11.2/4 m/s That is the max horizonall ball.

The initial velocity is now
vi=sqrt(4^2+(11.2/4)^2 )