(adsbygoogle = window.adsbygoogle || []).push({}); 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.

[tex]\theta[/tex]=73

[tex]\Delta[/tex]y=47m

[tex]\Delta[/tex]x=23m

2. Relevant equations

[tex]\Delta[/tex]y=v_{0y2}/2g

[tex]\Delta[/tex]x=v_{0x}t

v_{fx}=v_{0x}

[tex]\Delta[/tex]y=v_{0y}t-1/2gt^{2}

y=tan[tex]\theta[/tex]x-g/(2v_{0}^{2}cos^{2}[tex]\theta[/tex])*x^{2}

3. The attempt at a solution

47=v_{0y}^{2}/(2*9.81)

v_{0y}=[tex]\sqrt{}(2)(9.81)(47)[/tex]=30.37

30.37=v_{0}sin73

v_{0}=31.76

By using last equation, I got y=45.13, y coordinate when horizontal displacement is 23 meters.

v_{fy}=[tex]\sqrt{}(30.37)^{2}-(2*9.81*45.13)[/tex]=6.07

t=45.13

23=v_{0x}*45.13

v_{0x}=.51

v=6.09

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# Homework Help: Projectile motion

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