1. Part 2: projectiles launched at an angle Pages 102 – 104 Motion in Two Dimensions

2. Maximum height and range When objects are launched at an angle, there is a component of their velocity in the horizontal direction and a component in the vertical direction. The horizontal velocity would be v i cosθ The vertical velocity would be v i sinθ Problems can then be worked in the same way. Keep in mind that the vertical velocity at the peak of the path is zero.

3. Launched at an angle

5. Velocity Final speed = initial speed (conservation of energy) Impact angle = - launch angle (symmetry of parabola )

6. Launched at an angle The launch angle has an affect on the range of the projectile. As the angle increases from 0 to 45 degrees, the range increases. At 45 degrees, the range is at a maximum As the angle increases from 45 degrees to 90 degrees, the range decreases.

7. Launched at an angle

9. Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION a x = 0a y = -9.81 m/s 2 VELOCITY v x = v i cos θv yf = v i sin θ + aΔt v fy 2 = v i 2 sin θ +2a Δy DISPLACEMENT Δx = v i cos θ ΔtΔy = v i sinθΔt + ½ aΔt 2

10. The Hulk throws a boulder onto a police car trying to arrest him. The initial velocity of the boulder is 12 m/s and he throws it at an angle of 39 degrees to the horizontal: a. Find the horizontal and vertical components of the initial velocity b. Find the maximum height of the boulder c. Find the time the boulder was in the air d. Find the horizontal distance the boulder traveled

11. 39° vivi v i = 12m/s

12. 39° vivi Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy

13. Finding the components v xi = v i cos θ v yi = v i sin θ Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy

14. Find time first!!! v yf = v i sin θ + aΔt Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy

15. Finding max height Δy = v i sinθΔt + ½ aΔt 2 Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy

16. Finding the time in the air Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy

17. Finding the horizontal distance Δx = v i cos θ Δt Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy