Aprojectile is launched with an initial velocity of 25.0 m/s at an angle of 60o above the horizontal. at some time after its launch, the projectile’s velocity vector makes an angle of 30o with respect to the horizontal. what is the projectile’s speed at that point?

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.