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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.
[tex]\theta[/tex]=73
[tex]\Delta[/tex]y=47m
[tex]\Delta[/tex]x=23m
Homework Equations
[tex]\Delta[/tex]y=v0y2/2g
[tex]\Delta[/tex]x=v0xt
vfx=v0x
[tex]\Delta[/tex]y=v0yt-1/2gt2
y=tan[tex]\theta[/tex]x-g/(2v02cos2[tex]\theta[/tex])*x2
The Attempt at a Solution
47=v0y2/(2*9.81)
v0y=[tex]\sqrt{}(2)(9.81)(47)[/tex]=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=[tex]\sqrt{}(30.37)2-(2*9.81*45.13)[/tex]=6.07
t=45.13
23=v0x*45.13
v0x=.51
v=6.09
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