Answer:
A. [tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B. 5556.1 m
Explanation:
A.
Launch speed, vo
Angle of projection = θ
The value of vertical component of velocity at maximum height is zero. Let the maximum height is H.
Use third equation of motion in vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2a_{y}H[/tex]
[tex]0^{2}=\left (v_{0}Sin\theta \right )^{2}-2gH[/tex]
[tex]H=\frac{v_{0}^{2}Sin^{2}\theta }{2g}[/tex]
B.
u = 330 m/s
Let it goes upto height H.
V = 0 at maximum height
Use third equation of motion in vertical direction
[tex]v^{2}=u^{2}+2as[/tex]
[tex]0^{2}=330^{2}-2\times 9.8\times H[/tex]
H = 5556.1 m
