plus one projectile motion

1. PROJECTILE MOTION

3. Part 1. Motion of Objects Projected Horizontally

4. v 0 x y

5. x y

6. x y

7. x y

8. x y

11. Frame of reference: Equations of motion: x y 0 h v 0 y = h + ½ g t 2 x = v 0 t DSPL. v y = g t v x = v 0 VELC. a y = g = -9.81 m/s 2 a x = 0 ACCL. Y Accel. m. X Uniform m. g

12. Trajectory x = v 0 t y = h + ½ g t 2 Eliminate time, t t = x/v 0 y = h + ½ g (x/v 0 ) 2 y = h + ½ (g/v 0 2 ) x 2 y = ½ (g/v 0 2 ) x 2 + h Parabola, open down y x h v 01 v 02 > v 01

13. Total Time, Δt y = h + ½ g t 2 final y = 0 t i =0 t f = Δt 0 = h + ½ g ( Δt) 2 Solve for Δt: Total time of motion depends only on the initial height, h Δt = t f - t i y x h Δt = √ 2h/(-g) Δt = √ 2h/(9.81ms -2 )

14. Horizontal Range, Δx final y = 0, time is the total time Δt Horizontal range depends on the initial height, h, and the initial velocity, v 0 x = v 0 t Δ x = v 0 Δ t y x h Δt = √ 2h/(-g) Δx = v 0 √ 2h/(-g) Δx

15. VELOCITY v x = v 0 v y = g t Θ v v = √v x 2 + v y 2 = √v 0 2 +g 2 t 2 tg Θ = v y / v x = g t / v 0

16. FINAL VELOCITY v x = v 0 v y = g t Θ Θ is negative ( below the horizontal line) v v = √v x 2 + v y 2 v = √v 0 2 +g 2 (2h /(-g)) v = √ v 0 2 + 2h(-g) tg Θ = g Δt / v 0 = -(-g)√2h/(-g) / v 0 = - √2h(-g) / v 0 Δt = √ 2h/(-g)

17. HORIZONTAL THROW - Summary h – initial height, v 0 – initial horizontal velocity, g = -9.81m/s 2 v = √ v 0 2 + 2h(-g) tg Θ = -√2h(-g) / v 0 Final Velocity Δx = v 0 √ 2h/(-g) Horizontal Range Δt = √ 2h/(-g) Total time Half -parabola, open down Trajectory

18. Part 2. Motion of objects projected at an angle

19. Initial position: x = 0, y = 0 v i x y θ v ix v iy Initial velocity: vi = v i [ Θ] Velocity components: x- direction : v 0x = v 0 cos Θ y- direction : v 0y = v 0 sin Θ

22. Equations of motion: y = v 0 t sin Θ + ½ g t 2 x = v 0x t = v 0 t cos Θ x = v 0 t cos Θ DISPLACEMENT v y = v 0 sin Θ - g t v x = v 0x = v 0 cos Θ VELOCITY a y = g = -9.81 m/s 2 a x = 0 ACCELERATION Y Accelerated motion X Uniform motion

23. Trajectory x = v i t cos Θ y = v i t sin Θ + ½ g t 2 Eliminate time, t t = x/(v i cos Θ) y x Parabola, open down y = b x + a x 2

24. Total Time, Δt final height y = 0, after time interval Δt 0 = v 0 Δ t sin Θ + ½ g ( Δ t) 2 Solve for Δt: y = v 0 t sin Θ + ½ g t 2 t = 0 Δ t x 0 = v 0 sin Θ + ½ g Δ t Δ t = 2 v 0 sin Θ (-g)

25. Horizontal Range, Δx final y = 0, time is the total time Δt x = v 0 t cos Θ Δ x = v o Δ t cos Θ x Δx y 0 sin (2 Θ) = 2 sin Θ cos Θ Δt = 2 v 0 sin Θ (-g) Δ x = 2v o 2 sin Θ cos Θ (-g) Δ x = v o 2 sin (2 Θ) (-g)

27. Trajectory and horizontal range v i = 25 m/s

29. Maximum Height v y = v o sin Θ + g t y = v 0 t sin Θ + ½ g t 2 At maximum height v y = 0 0 = v i0 sin Θ + g t up t up = Δt/2 t up = v 0 sin Θ (-g) h max = v 0 t up sin Θ + ½ g t up 2 h max = v 0 2 sin 2 Θ/(-g) + ½ g(v i 2 sin 2 Θ)/g 2 h max = v i 2 sin 2 Θ 2(-g)

30. Projectile Motion – Final Equations (0,0) – initial position, v i = v i [ Θ] – initial velocity, g = -9.81m/s 2 2 v 0 sin Θ (-g) v 0 2 sin 2 Θ 2(-g) h max = Max height Δx = Horizontal range Δt = Total time Parabola, open down Trajectory v 0 2 sin (2 Θ) (-g)