Question

A projectile is launched over level ground at a launch angle of 70o with an initial velocity vo. At some later time while the projectile is on its way to the peak in its trajectory, its velocity vector makes an angle of 50o with respect to the horizontal. What is the magnitude of the projectile’s horizontal velocity at that point?

Answer 1

Horizontal component of velocity shall be
`v_(fx)=v_(o)cos(70^(o))`

Step-by-step explanation:

Since the given projectile motion is under the influence of gravity alone which acts in vertical direction only and hence the acceleration shall act in vertical direction only and correspondingly if air resistance is neglected the acceleration in the horizontal direction shall be zero.

For zero acceleration in the horizontal direction the velocity in horizontal direction shall not change.

Mathematically

`v_(ix)=v_(fx)`

We have initial horizontal velocity =
`v_(ix)=v_(o)cos(70^(o))`

Thus this shall remain constant throughout the course of the motion.