Question

For the ballistic missile aimed to achieve the maximum range of 9500 km, what is the maximum altitude reached in the trajectory

Answer 1

Explanation:

The range R of a projectile is given the equation

$R = \dfrac{v_0^2}{g}\sin{2\theta}$

The maximum range is achieved when $\theta = 45°$ so our equation reduces to

$R_{max} = \dfrac{v_0^2}{g}$

We can solve for the initial velocity $v_0$ as follows:

$v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}$

or

$v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}$

$\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}$

To find the maximum altitude H reached by the missile, we can use the equation

$v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy$

At its maximum height H, $v_y = 0$ so we can write

$0 = (v_0\sin{45°})^2 - 2gH$

or

$H = \dfrac{(v_0\sin{45°})^2}{2g}$

$\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}$

$\:\:\:\:\:\:= 2.4×10^6\:\text{m}$