1. Physics

2. PHS 5042-2 Kinematics & Momentum Projectiles Projectile: Moving object whose motion (trajectory, position, velocity) depends only on initial position, initial velocity and gravitational acceleration (g) Non-rectilinear motion

3. PHS 5042-2 Kinematics & Momentum Projectiles Horizontal motion (uniform rectilinear) Vertical motion (rectilinear, uniform acceleration) P t V t V t P t

4. PHS 5042-2 Kinematics & Momentum Projectiles Horizontal motion ( v = constant) Vertical motion (a = g) Initial velocity v x1 = v 1 cos θ v y1 = v 1 sin θ Initial positionx 1 = 0y1y1 Velocity (at time “t”)v x = v x1 = constantv y = g t + v y1 Position (at time “t”)x = (v x1 )(t)y = gt 2 /2 + (v y1 )(t) + y 1 y – y 1 = v y2 2 – v y1 2 /2g Velocity vector (at time “t”): V = √(v x 2 + v y 2 ) tan θ = v y / v x

5. PHS 5042-2 Kinematics & Momentum Kinematic Equations Example: You want to throw a stone 9.2 meters far from a window 3.25 meters high. What would be the horizontal speed needed? v x = d / t (horizontal motion) d = 9.2m; t = ? y = gt 2 /2 + (v y1 )(t) + y 1 (vertical motion) y 1 = 3.25m; v y1 = 0; y = 0 0 = (-9.8 m/s 2 )t 2 /2 + 3.25m - 3.25m = (- 4.9 m/s 2 ) t 2 t = √(3.25m) / (4.9 m/s 2 ) t = 0. 81 s v x = 9.2m / 0.81s v x = 11.35 m/s

6. PHS 5042-2 Kinematics & Momentum Kinematic Equations Example: A quarterback throws a touchdown pass from a height of 1.90 meters by throwing the ball at 60 degrees with a speed of 4 m/s. _What is the highest altitude of the ball? y = gt 2 /2 + (v y1 )(t) + y 1 g = - 9.8 m/s 2 ; v y1 = ?; y 1 = 1.9m; t = ? v y1 = v 1 sin θ v y1 = 4 m/s sin 60° v y1 = 4 m/s (0.87) v y1 = 3.48 m/s v y = g t + v y1 0 = g t + v y1 t = - v y1 / g = (-3.48 m/s) / (- 9.8 m/s 2 ) t = 0.355 s y = (- 4.9 m/s 2 ) (0.355 s) 2 + (3.48 m/s)(0.355 s) + 1.9m y = 2.52 m (highest altitude) *Try with a simpler equation (y – y 1 = v y2 2 – v y1 2 /2g)

7. PHS 5042-2 Kinematics & Momentum Kinematic Equations Example: A quarterback throws a touchdown pass from a height of 1.90 meters by throwing the ball at 60 degrees with a speed of 4 m/s. _How long will it stay in the air?

8. PHS 5042-2 Kinematics & Momentum Kinematic Equations Example: A quarterback throws a touchdown pass from a height of 1.90 meters by throwing the ball at 60 degrees with a speed of 4 m/s. _How far away will it land? x = (v x1 )(t) v x1 = v 1 cos θ v x1 = 4 m/s cos 60° v x1 = 4 m/s (0.5) v x1 = 2 m/s x = (2 m/s) (1.07s) x = 2.14 m (distance travelled)

9. PHS 5042-2 Kinematics & Momentum Kinematic Equations Practice Exercises: Page 5.45 – 5.49 Ex. 5.17 – 5.26