Velocity

[Guest]

A 10 lb. projectile with a diameter of 6 in. Find v at the moment 1 min. after leaving muzzle with velocity of 880 ft. per sec.

Not sure how to solve.

[Guest]

Neither the weight of the projectile nor the diameter of the muzzle is relevant. What is relevant is the angle at which the muzzle is aimed above the horizon.

If the angle is [tex]\theta[/tex] then the initial velocity is [tex]\left<880 cos(\theta), 880 sin(\theta)\right>[/tex] and the acceleration is [tex]\left< 0, -g\right>[/tex]. After t seconds, the velocity is [tex]\left<880 cos(\theta), 880 sin(\theta)- gt\right>[/tex]. After 1 minute, 60 seconds, the velocity is [tex]\left<880 cos(\theta), 880 sin(\theta)- 60(32.2)\right>[/tex] feet per second.

[Guest]

Thanks.