Physics
posted by Bill on .
A projectile is launched with an initial velocity of 268 m/s, at an angle of 45 degrees above the horizontal. At a certain point, A, in its motion, its velocity angle is 17 degrees above the horizontal. At another point, B, later in its motion, its velocity angle is 30 degrees below the horizontal. What is the horizontal distance from point A to point B?

This is a rather silly problem.
You know horizontal velocity is constant
horizontal velocity=268Cos45
So the angle changes because of vertical velocity. The resultant angle, in the problem then is a tangent question
Vertical velocity=268Sin459.8 t
tanTheta=vertical/horizontal
you know tan for 17, and 45, so the idea is to figure time for each angle from this e quation
horizontavelocity*tanTheta=268Sin459.8 t
Once you have the two times, then distance=horzontal velocity (t1t2)