A projectile is launched at an angle above the

ground. The horizontal component of the

projectile's velocity, vx, is initially 40. meters per

second. The vertical component of the projectile's

velocity, vy, is initially 30. meters per second.

What are the components of the projectile's

velocity after 2.0 seconds of flight? [Neglect

friction.]

(1) vx = 40. m/s and v

y = 10. m/s

(2) vx = 40. m/s and v

y = 30. m/s

(3) vx = 20. m/s and v

y = 10. m/s

(4) vx = 20. m/s and v

y = 30. m/s

The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u + at

I will take a to be - 9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)

v = 30 - 19.62

=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:

(1) vx=40m/s and vy=10m/s