Launching an Object at 73 Degrees: Velocity Calculation

[eestep]

Homework Statement

An object is launched from ground at an angle of 73 degrees from horizontal and reaches a maximum height of 47 meters. What is magnitude of object's velocity when object's horizontal displacement is 23 meters? Answer in m/s. Answer is 10.61.
\theta=73
\Deltay=47m
\Deltax=23m

Homework Equations

\Deltay=v_0y²/2g
\Deltax=v_0xt
v_fx=v_0x
\Deltay=v_0yt-1/2gt²
y=tan\thetax-g/(2v₀²cos²\theta)*x²

The Attempt at a Solution

47=v_0y²/(2*9.81)
v_0y=\sqrt{}(2)(9.81)(47)=30.37
30.37=v₀sin73
v₀=31.76
By using last equation, I got y=45.13, y coordinate when horizontal displacement is 23 meters.
v_fy=\sqrt{}(30.37)²-(2*9.81*45.13)=6.07
t=45.13
23=v_0x*45.13
v_0x=.51
v=6.09

 

[Delphi51]

I agree with your calc down to v0=31.76, but don't know where you got "y=45.13" or what it means. I didn't get the specified answer either, but maybe we can pool our work to find it!

initial horiz velocity = 31.76*cos(73) = 9.28
Time to x = 23: 23 = 9.28*t, so t = 2.477
vertical velocity at this time: Vy = Voy + at
Vy = 30.37-9.81*2.477 = 6.07
combined speed at time 2.477: v² = 9.28² + 6.07², v = 11.09

 

[JJBladester]

I understand a lot of times working with tags you may have to edit your post, but each time you do, you will see additional tags as below:<br /> <b><br /> 1. Homework Statement <br /> <br /> <br /> <br /> 2. Homework Equations <br /> <br /> <br /> <br /> 3. The Attempt at a Solution <br /> </b><br /> <br /> For some reason the forum adds them again each time you edit your post, so be careful to just remove them.

 

[eestep]

Answer calculated is agreeable, but I can't understand what's posted beneath it. Are you willing to explain it to me?

 

[JJBladester]

eestep, please clarify what you are need help with (write it out fully).

 

[eestep]

I appreciate all of your help and attempted it again successfully.