Answer:
v'=0.83m/a and v=10.2m/s
Explanation:
The information that we have is:
[tex]m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m[/tex]
The maximum height of the projectile is given by the equation
[tex]h=\frac{v^2sin^2\theta}{2g}[/tex]
So, rearrange for the velocity,
[tex]v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s[/tex]
Apply the conservation of momentum,
[tex]mvcos(\theta)=m'v'[/tex]
Then rearrange the recoil speed,
[tex]v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\[/tex]
