# Projectile motion

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

## Homework Equations

$$\Delta$$y=v0y2/2g
$$\Delta$$x=v0xt
vfx=v0x
$$\Delta$$y=v0yt-1/2gt2
y=tan$$\theta$$x-g/(2v02cos2$$\theta$$)*x2

## 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:

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Delphi51
Homework Helper
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

Gold Member
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. Homework Statement

2. Homework 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.

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