Given
Initial velocity is u=7.47 m/s
The angle is
[tex]\theta=81^0[/tex]The vertical height travelled, h=2.10 m
To find
The final speed of the pea.
Explanation
Along the vertical direction,
The final speed is given by the equation of kinematic
[tex]v_y^2=u_y^2+2ah[/tex]Putting the values
[tex]\begin{gathered} v_y^2=(7.47sin81^o)^2+2(-9.81)\times2.1 \\ \Rightarrow v_y=3.63\text{ m/s} \end{gathered}[/tex]Along the horizontal direction,
the speed is constant throughout the projectile
Thus,
[tex]v_x=7.47cos81^0=1.168\text{ m/s}[/tex]So the final velocity of the pea is
[tex]\begin{gathered} v=\sqrt{(3.63)^2+(1.168)^2} \\ \Rightarrow v=3.81\text{ m/s} \end{gathered}[/tex]Conclusion
The final velocity is 3.81 m/s
