Explanation:
The range R of a projectile is given the equation
[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]
The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to
[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]
We can solve for the initial velocity [tex]v_0[/tex] as follows:
[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]
or
[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]
[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]
To find the maximum altitude H reached by the missile, we can use the equation
[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]
At its maximum height H, [tex]v_y = 0[/tex] so we can write
[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]
or
[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]
[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]
