Answer to Question #263567 in Physics for Moi

Question

A volleyball is served overhead at an angle of 45 degrees above the horizontal at a speed of 9.8m/s. Show that the ball will just barely make it over the net and just about 2 meters on the other side of the net. Use projectile motion and free fall to calculate

a. The initial horizontal and vertical velocity 

b. The time it takes the ball to horizontally reach the net

Answer 1

a) $v_{0x}=9.8\cdot\cos45°=6.93\ (m/s)$

$v_{0y}=9.8\cdot\sin45°=6.93\ (m/s)$

b) $v_y=0=6.93-9.8\cdot t\to t=6.93/9.8=0.71\ (s)$

The height of the volleyball net is $2.43\ m.$

The ball reaches its maximum height $h_{max}=v_{0y}^2/(2g)=6.93^2/(2\cdot9.8)=2.45\ (m)$