The document discusses the motion of projectiles. It states that: 1) A projectile's velocity has two independent components: horizontal and vertical. The horizontal velocity does not change, while the vertical acceleration is constantly downwards at 9.81 m/s^2. 2) Examples show how to use kinematic equations to solve two-dimensional motion problems by separating the horizontal and vertical motions. 3) An example activity calculates that a ball thrown horizontally at 5 m/s from a roof 50 m high will hit the ground after 3.2 seconds and have traveled 16 meters horizontally.

1. The Motion of Projectiles
Chapter 11

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. 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. 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. 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. Activity
• Page 248, Q. 1-4
• Page 251, Q. 1- 5

7. Velocity at an Angle
•

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. 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. Activity
• Page 257, Q 1-4
• Page 261, Q.1-5