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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 + cos2 θ = 1; tan 2 θ + 1 = sec2 θ]
y = x tan θ - (½) g x2 / 2 v o2 cos2 θ
200 = 4000 tan θ - (9.81 × 40002 )/ (2 × 3002 cos2 θ)
200 - 4000 tan θ= - 872 sec2 θ= - 872 (1 + tan 2 θ)
∴ tan 2 θ - 4.587 tan θ + 1.23 = 0
∴ θ = 15.96o , or 76.91o
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