Please Help: The launching velocity of a projectile is 20m/s at 53 degrees above the horizontal. what is the vertical component of its velocity at launch? Neglecting air friction, which of these components remains constant throughout the flight path? which of these components determines the projectile's time in the air?

Question

anonymous · Answer

use trigonometry to get the vertical portion of the velocity:

20*sin(53) = Vy
20*cos(53) = Vx
Vx doesn't change (there is no horizontal force acting)
Vy determines how long it takes before gravity's acceleration brings it down.

anonymous · Answer

So 
the Vy is 20*sin(53)= 12.036 --> 12.04
the Vx is 20*cos(53)=15.97--> 16.00
right?

anonymous · Answer

Im not sure if this even right? 
im still studying the Physics: Projectile Motion.

anonymous · Answer

Vy=20sin35-gt=20sin35-9.81t
Vx=20cos35