Answer:
Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s
Explanation:
Given Data:
V is the overall velocity vector, [tex]Vi[/tex] and [tex]Ui[/tex] are its initial vertical and horizontal components
[tex]R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s[/tex]
To find:
Max Height [tex]h[/tex] achieved
Calculation:
1) Using the [tex]3^{rd}[/tex] equation of motion, we know
[tex]2*a*s = Vf^{2} - Vi^{2}[/tex]
2) In terms of gravity [tex]g[/tex] height [tex]h[/tex] and the vertical component of Velocity [tex]Vf , Vi[/tex].
3) As [tex]Vf = 0[/tex] as at maximum height the vertical component of velocity is zero maximum height achieved
[tex]2*g*h = Vf^{2} -Vi^{2}[/tex]
putting values
4) [tex]h = 0.784 m/s[/tex]
5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)
