Answer to Question #81049 in Classical Mechanics for Jinal Patel

Question

The shells fired from an artillery piece have a muzzle speed of 215 m/s, and the target is at a horizontal distance of 1.70 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

The horizontal distance that shell has traveled
L=(v_0^2 sin⁡2θ)/g
Therefore
sin⁡2θ=gL/(v_0^2 )=(9.81m/s^2×1700 m)/(〖215〗^2 m^2/s^2 )=0.36
2θ=(-1)^n arcsin⁡〖0.36+πn, n=0,±1,±2,…〗
2θ=(-1)^n 21.1° +180°n, n=0,±1,±2,…
θ=(-1)^n 10.5° +90°n, n=0,±1,±2,…
So
θ_1=10.5°,θ_2=-10.5°+90°=79.5°
Answer: 10.5°