Asked by Mea
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.
Answers
Answered by
jer
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]
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]
Answered by
Damon
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)
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)
Answered by
Mea
thank you so much
Answered by
Damon
Cool jer. Never saw it done like that before :)
Answered by
Damon
You are welcome.
Answered by
Mea
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