Projectile Motion in 2D: Kinetic Equations and Example Flight Analysis

1. Lesson 3: Physics 150 / 215Projectile Motion 2D Kinetic Equation of Motion Example of Projectile Flight.

2. 1-Dimensional Motion + x(t) - origin • + = positions to the right of the origin • - = positions to the left of the origin • x(t) = distance from origin at time t • = coordinate of position at time t

3. path = trajectory position x x(t) (meters) t time (seconds) Position - Time graph

4. Vector Kinetic Equations of Motion 1 ( ) ( ) ( ) r t = a t + v 0 t + r 0 2 2 1 ( ) ( ) º d t = a t + v 0 t 2 2 ( ) ( ) v t = a t + v 0 Kinetic Equations for each Û component / coordinate

5. ! Use the Kinetic Equations of motion separately on each component.

6. Projectile Motion in Gravity

7. V0=150m/s @ 30o 5m V0=150 cos30 i + 150 sin30 j

8. Initial Data • vertical direction • acceleration = -10m/s2 due to force of gravity • initial velocity = 150 sin 30 m/s • initial vertical position y=0 • horizontal direction • acceleration = 0 m/s2 • initial velocity = 150 cos 30 m/s • initial horizontal position x=0

9. At maximum height Vy=0 Therefore: 0 = -10t + 150 sin 30 thus t = (150 sin 30 / 10) s maximum height: h = (1/2)(-10) (150 sin 30 / 10)2 + (150 sin 30) (150 sin 30 / 10) = (1/2) (150 sin 30 / 10)2 m

10. Final vertical position: y =-5 Hence using displacement equation in vertical direction we can get time it takes to drop from max height to this vertical position -5- (1/2) (150 sin 30 / 10)2 = (1/2)(-10)t2drop Notice that initial vertical velocity is now 0 m/s as this is vel. At max. height.

11. Now we can find the range, which is the sum of the horizontal distance traveled while the object attains max. height (d1) and the horizontal distance traveled while the object drops from this height to y=-5 m, (d2). d1= (150 cos 30)tmax height d2 = (150 cos 30)tdrop and we have these times from the preceding slides.