Gun fires - projectile motion:
A gun fires a projectile along with a muzzle velocity of 300 m/sec as illustrated in Figure. Find its angle of inclination so that it strikes a target located at a horizontal distance of 4000 m and 200 m above it.
Solution
Coordinates (with respect to the origin O) of the target shall be (4000, 200). Substituting these values in the equation of motion, we attain
[sin^{ 2} θ + cos^{2} θ = 1; tan^{ 2} θ + 1 = sec^{2} θ]
y = x tan θ - (½) g x^{2} / 2 v _{o}^{2} cos^{2} θ
200 = 4000 tan θ - (9.81 × 4000^{2} )/ (2 × 300^{2} cos^{2} θ)
200 - 4000 tan θ= - 872 sec^{2} θ= - 872 (1 + tan ^{2} θ)
∴ tan ^{2 } θ - 4.587 tan θ + 1.23 = 0
∴ θ = 15.96^{o} , or 76.91^{o}