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

2. - - = positions to the left of the origin
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
1 5

3. x(t) Position - Time graph path = trajectory position x (meters) t
(seconds)
Position - Time graph
1 6

4. Vector Kinetic Equations of Motion1 ( ) ( ) ( ) r t = a t 2 + v t + r 2 1 ( ) ( ) d t = a t 2 + v t 2 ( ) ( ) v t = a t + v
Kinetic Equations for each Û
2 9
component /
coordinate

5. ! Use the Kinetic Equations of motion separately on each component.
3 0

6. Projectile Motion in Gravity34

7. 30o 5m
V0=150 cos30 i sin30 j

8. acceleration = -10m/s2 due to force of gravity
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 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. and we have these times from the preceding slides.
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.