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# Physics 504 Chapter 11 The Motion of Projectiles

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### Physics 504 Chapter 11 The Motion of Projectiles

1. 1. The Motion of Projectiles Chapter 11
2. 2. Velocity Components• The velocity of a projectile is comprised of two independent components: horizontal and vertical.• Time of flight is determined by vertical motion only.• Vertical acceleration is downwards at 9.81 m/s2• Horizontal velocity does not change• Horizontal acceleration = 0
3. 3. Solving 2D Motion Questions• Vertical motion uses one of the 3 equations• v = u + at• s = ut + ½ at2• v2 = u2 + 2as• Horizontally: Distance = velocity x time
4. 4. Activity• From a roof of a building 50 m high, a ball is thrown horizontally with a v of 5 m/s. When does the ball hit the ground and at what distance?• Given: Vertically: s = 50 m, u = 0, a = 9.81 m/s2• Given: Horizontally: v = 5 m/s• Calculate time:• Calculate distance:
5. 5. Activity• Time: s = ut + ½ at2• 50m = 0t + ½ 9.81t2• t = 3.2 s• Distance: d = vt• = 5 m/s x 3.2 s• = 16 m
6. 6. Activity• Page 248, Q. 1-4• Page 251, Q. 1- 5
7. 7. Velocity at an Angle•
8. 8. Example• A soccer player kicks a ball at 20 m/s at 25°• Time to teach max height:• Initial vx = v cosθ = 20 cos 25° = 18.1m/s• Initial vy = v sinθ = 20 sin 25° = 8.45 m/s• v = uy + at; 0 = 8.45 -9.81t; t = 0.862 s• Time of flight: time up = time down• 2 x 0.862 s = 1.724 s
9. 9. Continued• Horizontal distance:• d =vxt = 18.1m/s x 1.724s = 31.2 m• Velocity and angle at contact:• v = √ (vx2 + vy2); v = u + at = 0 + 9.81x 0.762 = 8.45 m/s y• = √ (18.12 + 8.452)• = 20 m/s• Angle of contact = tan -1(8.45/18.1)=0.467• = 25° = 335°
10. 10. Activity• Page 257, Q 1-4• Page 261, Q.1-5