A volleyball court is 18 m long from baseline to baseline and the net is 2.24 m high. A volleyball player serves the ball from the baseline by hitting it at a height of 2 m above the floor at an angle of 10 degrees above the horizontal.

Question

A.) If the ball is in the air for 1.05 s, is the serve in or out (assuming it clears the net)?

B.) Does the serve actually clear the net?

Answer 1

It goes up, then down for 1.05 s
do the vertical problem

Hi = 2
final h = 0

h = Hi + Vi t - 4.9 t^2
0 = 2 + 1.05 Vi - 4.9 (1.05)^2
-2 + 5.4 = 1.05 Vi
Vi = 3.24 m/s initial speed up

Now the horizontal problem
max speed * sin 10 = 3.24
so speed = 18.7 m/s
so
u = 18.7 cos 10 = 18.4 m/s
distance = u t = 19.3 meters
so out

time at 9 meters
t = 9/18.4 = .489 seconds to net
h = 2 + 3.24(.489) - 4.9(.489)^2
= 2.41 so it clears the net

Answer 2

Thank you so much!!!!!! :)