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?

A. 0

B. g

C. vo*cos50

D. vo*sin50

E. vo*cos70

Answer 1

The magnitude of the projectile's horizontal velocity at the point where its velocity vector makes an angle of 50° with respect to the horizontal is vo ∙ cos 70°.

Step-by-step explanation:

The magnitude of the projectile's horizontal velocity at the point where its velocity vector makes an angle of 50° with respect to the horizontal can be found using the equation:

Vx = vo * cos θ

Where Vx is the horizontal velocity, vo is the initial velocity, and θ is the launch angle. In this case, the launch angle is 70° and the initial velocity is not given. Therefore, the magnitude of the projectile's horizontal velocity at that point is option E, vo ∙ cos 70°.