Homework Help: Projectile motion

1. Sep 26, 2010

eestep

1. The problem statement, all variables and given/known data

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
$$\Delta$$y=47m
$$\Delta$$x=23m

2. Relevant equations
$$\Delta$$y=v0y2/2g
$$\Delta$$x=v0xt
vfx=v0x
$$\Delta$$y=v0yt-1/2gt2
y=tan$$\theta$$x-g/(2v02cos2$$\theta$$)*x2
3. The attempt at a solution
47=v0y2/(2*9.81)
v0y=$$\sqrt{}(2)(9.81)(47)$$=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.
vfy=$$\sqrt{}(30.37)2-(2*9.81*45.13)$$=6.07
t=45.13
23=v0x*45.13
v0x=.51
v=6.09

Last edited: Sep 27, 2010
2. Sep 26, 2010

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

3. Sep 26, 2010

I understand a lot of times working with [tex] tags you may have to edit your post, but each time you do, you will see additional tags as below:

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

For some reason the forum adds them again each time you edit your post, so be careful to just remove them.

4. Sep 27, 2010

eestep

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

5. Sep 28, 2010