What is the vertical and horizontal components of veloctiy?

[Hawke]

The launching velocity of a projectile is 20 m/s at 53 degrees above the horizontal. What is the vertical component of its velocity at launch? Its horizontal component of velocity? Neglecting air friction, which of these components remains constant though out the flight path? Which of these components determines the projectile's time in the air?

[Hungry!]

vertical component would be sin theta and horizontal component would be cos thetha

[j_sun]

Let U = 20 mps and theta = 53 deg re the horizontal.

Then Uy = U sin(theta) = 20*sin(53) is the vertical and Ux = U cos(theta) = 20*cos(53) the horizontal.

Discounting drag forces, the only...only...force acting on the object is the force of gravity W = mg.  And W is strictly vertical as that's the way gravity works, up and down.  As there is no...no...horizonta l force, we have Fx = mAx = 0 and Ax, the horizontal accleration, = 0.  In which case Vx = Ux + AxT = Ux + 0T = Ux.  In other words, the horizontal speed remains at Ux; so it's constant.

But there is a vertical force, gravity.  So Fy = mAy = mg <> 0; so that the acceleration due to gravity g <> 0.  In which case, Vy = Uy - gT and (- Vy + Uy)/-g = T determines how long the projectile remains in the air.  So both the initial Uy and final Vy vertical speed components do this.

NOTE - g, which means the gravity force is decelerating the initial speed Uy which is a + value and assumed upward.  But as time wears on, Vf will eventually turn negative, meaning that the end speed, Vf, will most likely be negative and in the direction of the gravity force that always...always...a cts downward.