Physics Help Forum Projectile motion: Missile over a hill
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 Jun 3rd 2019, 01:03 AM #1 Junior Member   Join Date: Jun 2019 Posts: 2 Projectile motion: Missile over a hill 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?
Jun 4th 2019, 10:39 AM   #2
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Join Date: Jan 2019
Posts: 20
 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 $v_0$

 Jun 4th 2019, 06:50 PM #3 Junior Member   Join Date: Jun 2019 Posts: 2 Thank you for replying! Sorry I didn't include my attempt along with the post😔

 Tags hill, missile, motion, projectile, wall

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