Question

The shells fired from an artillery piece have a muzzle speed of 225 m/s, and the target is at a horizontal distance of 1.20 km. two possible angles can be used to hit the target. find the smallest angle relative to the horizontal at which the gun should be aimed?(hint: sin2θ = 2sinθcosθ)

Answer 1

Given

The parametric equations for shell position are ...

... (x, y) = (v₀·cos(θ)t, -(1/2g)t²+v₀·sin(θ)t)

with conditions

... v₀ = 225 m/s

... g = 9.8 m/s²

Find

The value of θ in the range 0–45° that makes (x, y) = (1200, 0) m for t > 0.

Solution

Solving the x equation for t, we have

$x=v_0\cos{(\theta)}t=d\\\\t=\dfrac{d}{v_0\cos{(\theta)}}$

Substituting this into the equation for y gives

$y=-\dfrac{g}{2}t^2+v_0\sin{(\theta)}t=0\\\\v_0\sin{(\theta)}-\dfrac{g}{2}t=0\qquad\text{divide by t}\\\\\dfrac{2v_0\sin{(\theta)}}{g}=\dfrac{d}{v_0\cos{(\theta)}}\qquad\text{solve for t, substitute for t from above}\\\\v_0^2\sin{(2\theta)}=gd\\\\\theta=\dfrac{1}{2}\arcsin{\left(\dfrac{9.8\cdot 1200}{225^2}\right)}\approx 6.72^{\circ}$