Gun fires - projectile motion, Mechanical Engineering

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.

1430_Gun fires - projectile motion.jpg

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

Posted Date: 1/29/2013 1:59:42 AM | Location : United States







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