Originally Posted by **Leonson** A missile is to be launched from a point P, 500m from the peak of a hill of height 2000m into enemy territory. If the missile is to just pass over the hill and strike a target 1200m from P, what is the velocity at which the missile should be launched, and the angle of launch? |

$x = v_0 \cos{\theta} \cdot t \implies t = \dfrac{x}{v_0 \cos{\theta}}$

$y = v_0 \sin{\theta} \cdot t - \dfrac{g}{2}t^2$

eliminating the parameter of time yields the equation

$\color{red}{y = x\tan{\theta} - \dfrac{gx^2}{2v_0^2\cos^2{\theta}}}$

using position coordinates $(x,y)$, in the general quadratic $y=ax^2+bx+c$

$P=(0,0)$, $(500,2000)$, and $(1200,0)$

$0 = a\cdot 0^2 + b \cdot 0 + c \implies c = 0$

$2000 = a \cdot 500^2 + b \cdot 500 \implies 500a+b=4$

$0 = a \cdot 1200^2 + b \cdot 1200 \implies 1200a + b = 0$

solving the system of equations yields

$\color{red}{y = -\dfrac{1}{175}x^2 + \dfrac{48}{7}x}$

equating the linear coefficient:

$\tan{\theta} = \dfrac{48}{7} \implies \theta = \tan^{-1}\left(\dfrac{48}{7}\right) \text{ and } \cos{\theta} = \dfrac{7}{\sqrt{48^2+7^2}}$

equating the quadratic coefficient:

$\dfrac{g}{2v_0^2 \cos^2{\theta}} = \dfrac{1}{175}$

you can solve for the magnitude of $v_0$