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.
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]
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)
Cool jer. Never saw it done like that before :)
You are welcome.
thank you so much
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
To calculate the speed of the ball just before the opposing player strikes it, we can use the principle of conservation of mechanical energy.
Here's how you can solve the problem step by step:
1. Break down the initial velocity of the ball into its horizontal and vertical components. The horizontal component will remain constant throughout the trajectory, while the vertical component will change due to the effects of gravity.
The initial horizontal velocity is given to be 8.5 m/s, and the angle is 32 degrees above the horizontal. Therefore, the initial horizontal component of velocity can be calculated as:
Initial horizontal velocity (Vx) = 8.5 m/s * cos(32 degrees).
The initial vertical velocity can be calculated as:
Initial vertical velocity (Vy) = 8.5 m/s * sin(32 degrees).
2. Determine the time it takes for the ball to reach the opposing player's height by using kinematic equations for vertical motion. We know that the initial vertical displacement (Δy) is 2.36 m and the initial vertical velocity (Vy) is given. Assuming the player hits the ball at the highest point, you can use the following kinematic equation:
Δy = Vy * t - (1/2) * g * t^2,
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearrange the equation to solve for time (t), giving:
t = (√(Vy^2 + 2 * g * Δy) - Vy) / g.
3. Once you have the time it takes for the ball to reach the opposing player's height, you can calculate the final horizontal displacement (Δx) using the formula:
Horizontal displacement (Δx) = Vx * t.
4. Calculate the final vertical displacement (Δy) from the initial height (1.4 m) to the opposing player's height (2.36 m) using the formula:
Vertical displacement (Δy) = final height - initial height.
5. Use the Pythagorean theorem to calculate the magnitude of the final velocity just before the opposing player strikes the ball. The final velocity (Vf) can be calculated as:
Vf = √(Δx^2 + Δy^2).
By following these steps, you should be able to determine the speed of the ball just before the opposing player strikes it.