- #1

- 36

- 0

## 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=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}

## 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

Last edited: