# Finding the final velocity of a volleyball

• benca
In summary, the conversation discusses calculating the y component of the initial velocity vector and the change in distance for a volleyball player serving a ball. The problem arises when attempting to solve for the final velocity using the equation v2y^2 = v1y^2 + 2aydy, leading to a discrepancy between the calculated and given heights. This suggests a possible error in the problem's numbers.
benca
Homework Statement
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 it 3.6 m above the court, returning it over the net. Calculate the speed of the ball just before the opposing player strikes it
Relevant Equations
v2y^2 = v1y^2 + 2aydy
First I calculated the y component of the initial velocity vector:

vy1 = 8.5 m/s * sin32
= 4.7 m/s

next the change in distance

Δd = d2 - d1
= 3.6m - 1.4m
= 2.2m

Then I put these numbers into the equation v2y^2 = v1y^2 + 2aydy

v2y^2 = (4.7 m/2)^2 + 2(-9.8 m/s^2)(2.2m)

The problem is I end up needing to take the square root of a negative number to solve for v2y^2. I'm not really sure what I'm doing wrong. ay should be negative because it's in the downward/negative direction. I thought the problem might be something to do with the way I'm thinking about Δd, but I don't know where to go from here.

benca said:
Homework Statement: 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 it 3.6 m above the court, returning it over the net. Calculate the speed of the ball just before the opposing player strikes it
Homework Equations: v2y^2 = v1y^2 + 2aydy

First I calculated the y component of the initial velocity vector:

vy1 = 8.5 m/s * sin32
= 4.7 m/s

Okay, so far. Why not calculate the maximum height of the ball? Just to see. It might be interesting

Alright, I used v2y^2 = v1y^2 + 2aydy, set v2y^2 to 0 and solved for dy.

dy = (v2y^2 - v1y^2)/2ay

dy = (0 - 4.7m/s)/(2*-9.8m/2^2)

dy = 1.1m

then, I solved for d2y:

Δdy = d2y - d1y
1.1m = d2y - 1.4m
d2y = 3.5m

That seems suspiciously close to the height of 3.6m given in the problem. If it is supposed to be 3.6 and I didn't round or calculate something properly, then that would give me the final velocity in the y direction.

After that, calculate the initial velocity in the x direction using 8.5*cos(32). Since there's no mention of external force acting in the x direction, that means the vx2 = vx1 ? Then use pythagoreans theorem to solve for the final speed?

Although I'm not sure how to account for the difference in the height I calculated and the given height of 3.6m

edit* the height I got was 2.5 zzz . anyways, so that definitely means I got something wrong. still unsure of what though

Last edited:
benca said:
Alright, I used v2y^2 = v1y^2 + 2aydy, set v2y^2 to 0 and solved for dy.

dy = (v2y^2 - v1y^2)/2ay

dy = (0 - 4.7m/s)/(2*-9.8m/2^2)

dy = 1.1m

Well, exactly. The ball is only going to gain 1.1m and is not going to get to 3.6m.

Physics, believe it or not, is supposed to be about the real world. It's not just a set of abstract equations.

The first thing I thought when I looked at the question was that 3.6m is very high. Even if the player is 2m tall that is a long way up. And I couldn't see how the ball or the player was going to get that high.

All your work is correct. Except 1.4 + 1.1 = 2.5 and not = 3.5.

In any case, you should see from the work you have done that the numbers in the problem are wrong.

ah ok, I see. I'll try to keep the possibility of errors in mind as I go through this course. Thanks

benca said:
ah ok, I see. I'll try to keep the possibility of errors in mind as I go through this course. Thanks
For example, a basketball hoop is just over 3m high. So, 3.6m is too high to hit a volleyball!

## 1. How is the final velocity of a volleyball calculated?

The final velocity of a volleyball can be calculated by using the formula vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.

## 2. What is the initial velocity of a volleyball?

The initial velocity of a volleyball is the speed at which it is launched or thrown. This can be measured using a radar gun or by timing how long it takes for the volleyball to travel a known distance.

## 3. What is the acceleration of a volleyball?

The acceleration of a volleyball is typically due to gravity and is constant at 9.8 meters per second squared. However, other factors such as air resistance can also affect the acceleration of a volleyball.

## 4. Does the weight of the volleyball affect its final velocity?

Yes, the weight of the volleyball can affect its final velocity. A heavier volleyball will require more force to move and will have a lower final velocity compared to a lighter volleyball with the same initial velocity and acceleration.

## 5. Can the final velocity of a volleyball be greater than its initial velocity?

Yes, the final velocity of a volleyball can be greater than its initial velocity if it is thrown or hit with enough force. This can also be affected by factors such as air resistance and the surface it is traveling on.

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