- #1

benca

- 19

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- 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.

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.