Atennis player hits a ball 2.0m above the ground. the ball leaves his racquet with a speed of 20m/s at an angle of 5.0 degrees above the horizontal. the horizontal distance to the net is 7.0m and the net is 1.0m high. does the ball clear the net? if so, by how much? if not, by how much does it miss.

Explanation:

Consider the horizontal motion of ball

    We have equation of motion s = ut + 0.5at²

    Initial velocity, u = 20 cos 5 = 19.92 m/s

    Acceleration, a = 0 m/s²

    Displacement,s = 7 m

Substituting

    s = ut + 0.5at²

    7 = 19.92 x t + 0.5 x 0 x t²

     t = 0.35 seconds

After 0.35 seconds ball reaches at net place.

Consider the vertical motion of ball

    We have equation of motion s = ut + 0.5at²

    Initial velocity, u = 20 sin 5 = 1.74 m/s

    Acceleration, a = -9.81 m/s²

    Time, t = 0.35 s

Substituting

    s = ut + 0.5at²

    s = 1.74 x 0.35 + 0.5 x -9.81 x 0.35²

    s = 0.008 m

Displacement with respect to ground = 2+0.008 = 2.008 m

Height of net  = 1 m

So the ball clears the net.

Height of clearance = 2.008 - 1 = 1.008 m  

explanation:

v=uf/(u-f)

= 5.-10/(5+10) convex lens has negative focal length

the answer would be   dignified - apex