What is the speed of the ball when it leaves Sarah's hand?

[Alcubierre]

Homework Statement

The question I am struggling with is the second version of a similar problem so to make it easier for me to receive assistance, I'll post both parts:

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 20 m/s at an angle 49 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

Now for the part I am struggling with:

After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 15 m above the ground, and takes 2.692 s to get directly over Julie's head. What is the speed of the ball when it leaves Sarah's hand?

Given from second part:

$t = 2.692 s$

$y_{0} = 1.5 m$

$y_{max} = 15 m$

$v_{0,max} = 0 \frac{m}{s}$

Homework Equations

Projectile motion equations

The Attempt at a Solution

So this problem is asking for the initial speed of the ball when it leaves Sarah's hand. I tried three different approaches:

First approach: I used the range (40.337 m) found in a previous question and used the horizontal displacement equation

$x = [v_{0}cos({\theta})]t$​

and solved for $v_{0}$ which gave me the wrong initial speed of 22.86 m/s.

Second approach: I used the vertical velocity equation

$v_{y} = v_{0}sin({\theta}) - gt$​

And solved for $v_{0}$ which gave me 34.99 m/s, which was also wrong.

My third and last approach, was using the same steps as above, but I later (wrongly) realized that I could set it equal to 0 because the velocity is 0 at $y_{max}$ and used half of the total time of flight, and that gave me a initial speed of 17.49 m/s, which also turned out to be wrong.

After I finished writing this post, I realized that maybe I am wrong to assume that Sarah is throwing the ball back at a 49 degree angle to the horizontal like Julie. I have 2 more tries before I get the question wrong (it's an online system) and I would really appreciate if you guys could lead me in the right direction.

Thank you,

Rafael

 

[Haye]

Yes, you indeed made a wrong assumption on the angle.
If you take a good look at the values given, you should be able to calculate v_x without much trouble.
Then you only need to calculate v_y.

 

[Alcubierre]

So would I use 45 degrees, disregard the angle altogether, or find the angle?

 

[Haye]

There is no angle given, so you shouldn't assume any angle. You can calculate v_x and v_y without using the angle. If you want to know the angle afterwards, you can calculate it from v_x and v_y.

 

[Alcubierre]

Okay this is what I did:

We know that the horizontal speed remains constant, so

$x = v_{0,x}t$

$v_{0,x} = \frac{x}{t} = \frac{40.377 m}{2.692 s} = 14.99 \frac{m}{s}$​

Then for the vertical component,

$v^2_{y} = v^2_{0,y} - 2g \Delta y$

$v_{0,y} = \sqrt{2gΔy} = 17.15 \frac{m}{s}$​

So the speed is,

$\left|v_{0} \right| = \sqrt{v^2_{0,x} + v^2_{0,y}} = 22.78 \frac{m}{s}$​

Is this correct?

 

[Alcubierre]

It is! I checked, thank you very much, Haye.