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