Finding the final velocity of a volleyball

[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.

 

[PeroK]

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

 

[benca]

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

 

[PeroK]

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.

 

[benca]

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

 

[PeroK]

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!