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 = 15.96^o ,   or   76.91^o