PROJECTILE MOTIONLaunched at an angle

Recall Definition of terms

› Range› Time of Flight› Maximum Height› Trajectory

Representation of Angular Projection

With Respect to Air Resistance

The trajectory of a high speed projectile falls short of a parabolic path.

Angular Projectile

To solve initial velocity:Vix = Vi cos Θ

Viy = Vi sin Θ

To solve for vertical velocityVfy = Viy + gt

For maximum height:(Vi sin Θ)2

2g

To solve for tt = Vi sin Θ

gt’ = 2Vi sin Θ

g

RangeVi 2 sin 2Θ g

Sample problemA long jumper leaves the ground at an

angle 30° to the horizontal and at a speed of 6m/s. How far does he jump?

Given Θ = 30° Vi = 6m/sFind: dx or R

Answert’ = 0.61sR = 3.20mdx = 3.18m