A volleyball player serves a ball, giving it an initial velocity of 8.5 m/s [32 up] at an initial height of 1.4 m above the court floor. An opposing player jumps to meet the ball and hits 2.36 m above the court, returning it over the net. Calculate the speed of the ball just before the opposing player strikes it.

Question

A volleyball player serves a ball, giving it an initial velocity of 8.5 m/s [32 up] at an initial height of 1.4 m above the court floor.  An opposing player jumps to meet the ball and hits 2.36 m above the court, returning it over the net.  Calculate the speed of the ball just before the opposing player strikes it.

jer · Accepted Answer

First we have to find time, and we can do that by finding Vx and d in the equation Vx = d/t

We find Vx with trig
cosX*V = Vx
cos32*8.5 = Vx
Vx = 7.2 m/s [right]

Then we find d also through trig
tanX = o/a
o/tanX = a
a = 0.96m/tan32
a = 1.54m

Then find t
Vx = d/t
d / Vx = t
(1.54m) / (7.2m/s [right]) = t
0.21s = t

Then we find Vyf, and to do that we must first find Viy, which can also be found through trig
sinX = o/h
sinX * h = o
sinX * Vi = Vyi
sin32 * 8.5m/s = Vyi
Vyi = 4.50m/s [up]

Then we can find Vyf through one of the kinematics
v2 = v1 + at
v2y = v1y + at
v2y = 4.5m/s + (-9.8m/s^2)(0.21s)
Vyf = 2.44m/s [up]

Now that we have Vx and Vyf, we can find the hypoteneuse of this which will be our Vf.

Rearrange a^2 + b^2 = c^2:
Vf = sqrt[ (vx^2) + (vyf^2) ]
Vf = sqrt[ (7.2^2 + 2.44^2)
Vf = 7.6m/s

Then you find angle also through trig
tanX = o/a
tanX = 2.44/7.2
tan-1(2.44/7.2) = X
X = 19 degrees

Answer:
Vf = 7.6 m/s [E 19 N]

Damon · Answer

Vi = initial vertical velocity = 8.5 sin 32 = 4.50 m/s
Hi = initial height = 1.4
g - 9.81 m/s^2
when will it be at 2.36 meters (once going up and once going down)?
h = Hi + Vi t - 4.9 t^2
2.36 = 1.4 + 4.5 t - 4.9 t^2
4.9 t^2 - 4.5 t + 0.96 = 0
https://www.mathsisfun.com/quadratic-equation-solver.html
t = 0.337 on the way up and 0.581 on the way down
Horizontal speed = u = 8.5 cos 32 = 7.21 meters/second
the on the way down time is too far away (7.21 m/s * .6 s = 4.3 meters)
I suspect he whacked it on the way up (7.21 * 0.337 = 2.43 meters)
so
v vertical = 4.50 - 9.81*.337 = 1.19 m/s still going up
u horizontal = 7.21 forever
so sqrt (1.19^2 + 7.21^2)

Mea · Answer

thank you so much

Damon · Answer

Cool jer. Never saw it done like that before :)

Damon · Answer

You are welcome.

Mea · Answer

Hey guys, I was trying to figure out how to solve it using energy equation Et= Ek1+Ek2, but I am finding the answer to be 7.3 instead of 7.6