Question

A volleyball player spikes the ball. The ball hits the floor with a velocity of 21.2 m/s at 30 degrees above the horizontal floor, 4.66 m away from where she spiked it. What was the ball's initial velocity (immediately after the player hits the ball)? Give a magnitude and sketch the direction.

first I used the equation
Vfinal ^2 - Vinitial^2= 2(xfinal-xinitial)
so...
(21.1 m/s)^2 - Vinitial^2 = 2(9.8m/s)(4.66-0)

445.21- Vinitial = (-19.6)(-4.66)
445.21- Vinitial = -91.336
Vinitial = 23.16 m/s

Is this how I would figure it out becuase I'm kind of lost

Answer 1

That cant be right. A ball gains velocity going down.

Figure the <b>vertical</b> velocity at the floor: 21.1*sin30
Now, that is vfinal. Solve for vintialy here
vfinaly= vinitialy+ 2*9.8*4.66

That will give you the initial vertical velocity. The intial horiontal velocity will be the same as the final horizontal velocty, 21.2 cos30.

Now, you can find the intial velocity.

Answer 2

Thank you very much for the help. So i figured out that the vertical initial velocity was 4.47 m/s and the horizontal initial velocity is the same as the final velocity, so it would be 18.27 m/s. So do I need to use both the vertical and horizontal velocties to find the initial velocity or do I just need to worry about the vertical velocity..would I add them? Im sorry, I'm confused