A vector has magnitude and direction. Choose axes and positive directions before resolving it. For speed u at angle θ above horizontal, ux=ucosθ and uy=usinθ. Recombine perpendicular components with v=√vx2+vy2. Use component signs to select the direction quadrant; negative vy means downward if upward is positive.
Once an ideal projectile leaves the launcher, gravity is its only force. There is no continuing forward launch force. With air resistance neglected, ax=0 and ay=−g throughout flight, including the highest point. Equal-time positions therefore have equal horizontal spacings but changing vertical spacings.
Example 1
A ball leaves at 20.0ms−1,30.0∘ above horizontal. Find its components.
ux=20.0cos30.0∘=17.3ms−1; uy=20.0sin30.0∘=10.0ms−1 upward.
Example 2
A projectile has vx=12.0ms−1 and vy=−5.00ms−1. Find its velocity.
v=√122+52=13.0ms−1. arctan(125)=22.6∘, so the direction is 22.6∘ below horizontal.
Watch out: With drag neglected, gravity points downward throughout flight; velocity can point upward, sideways or downward.