Projectile motion related problem

[Thewindyfan]

Homework Statement

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 18 m/s at an angle 39 degrees with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

5) After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground. It is moving with a speed of 15 m/s when it reaches a maximum height of 9 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?

Homework Equations

y = 1/2at^2 + Vo(y)t^2 + y(o)
V = V(o) + at

The Attempt at a Solution

Again, I have come across a problem where I know exactly how to do it, yet I'm still messing up somewhere that I am not sure of algebraically and would like to know where I'm going wrong in my approach to this question:
To find the initial velocity, we need both the horizontal component and vertical component of velocity. We are given the horizontal component of velocity to be 15 m/s, because when the ball is at max height the only velocity it has is its horizontal velocity. So we first need to find the vertical velocity before applying Pythagorean's theorem to find the resultant velocity vector.
- I do know there is the easy way of doing this without concerning time at all using the expression with Vf^2 - V(o)^2 and what not, but it's annoying me that I'm messing up somewhere in the problem, even though the algebra doesn't seem that difficult at all. So this is what I was doing -
Using the expression for velocity, I found that the time that it reaches the max height is Voy/a and then used it in the displacement equation:
9 = 1/2*a*(Voy^2/a^2) + Voy*(Voy/a) + y(o)
9 - y(o) = 1/2*(Voy^2/a) + Voy^2/a
9 - y(o) = (Voy^2/a)(1/2 + 1)
(9-y(o))/(3/2) = Voy^2/a
(9-y(o))*(2/3)*(a) = Voy^2
Voy = √(9-y(o))*(2/3)*(a)
When I arrived at this conclusion, I knew I messed up because when you plug in the numbers, the number inside the radical ends up being negative so it doesn't make sense. Can someone point out where I'm going wrong in my train of thought here? Thank you so much.

*I also retried the algebra earlier using a = -g, and still arrived at the same conclusion.

 

[cnh1995]

After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground. It is moving with a speed of 15 m/s when it reaches a maximum height of 9 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?

You have v_y=0, h=9m. Calculate initial vertical velocity from this. You have horizontal velocity too. Then calculate total velocity.

 

[Thewindyfan]

What will be the vertical velocity when the ball will reach maximum height?

Oh wow, I can't believe I made the mistake of leaving Voy*t in the expression. Now it makes sense!

Thanks for putting up with my block of text and getting straight to the point! I'll retry the problem later when I get back to it armed with this correction.

 

[cnh1995]

Oh wow, I can't believe I made the mistake of leaving Voy*t in the expression. Now it makes sense!

Thanks for putting up with my block of text and getting straight to the point! I'll retry the problem later when I get back to it armed with this correction.

Well, I thought you had considered that, so I edited the post! Good luck!