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Physics

Physics. 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. PHS 5042-2 Kinematics & Momentum Projectiles.

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Physics

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  1. Physics

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

  4. PHS 5042-2 Kinematics & MomentumProjectiles Velocity vector (at time “t”): V = √(vx2 + vy2) tan θ = vy / vx

  5. PHS 5042-2 Kinematics & MomentumKinematic 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? vx = d / t (horizontal motion) d = 9.2m; t = ? y = gt2/2 + (vy1)(t) + y1(vertical motion) y1 = 3.25m; vy1 = 0; y = 0 0 = (-9.8 m/s2)t2/2 + 3.25m - 3.25m = (- 4.9 m/s2) t2 t = √(3.25m) / (4.9 m/s2) t = 0. 81 s vx = 9.2m / 0.81s vx = 11.35 m/s

  6. PHS 5042-2 Kinematics & MomentumKinematic 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 = gt2/2 + (vy1)(t) + y1 g = - 9.8 m/s2; vy1 = ?; y1 = 1.9m; t = ? vy1= v1 sin θ vy1 = 4 m/s sin 60° vy1= 4 m/s (0.87) vy1 = 3.48 m/s vy= g t +vy1 0 = g t +vy1 t = - vy1 / g = (-3.48 m/s) / (- 9.8 m/s2) t = 0.355 s y = (- 4.9 m/s2) (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 – y1 = vy22 – vy12 /2g)

  7. PHS 5042-2 Kinematics & MomentumKinematic 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? y = gt2/2 + (vy1)(t) + y1 g = - 9.8 m/s2; vy1 = 3.48 m/s; y1 = 1.9m; t = ? 0 = (-9.8 m/s2)t2/2 + (3.48 m/s)t + 1.9m Solving with quadratic equation: t =[- (3.48 m/s) - ] / - 9.8 m/s2 t = 1.07 s (time in the air)

  8. PHS 5042-2 Kinematics & MomentumKinematic 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 = (vx1)(t) vx1= v1 cos θ vx1= 4 m/s cos 60° vx1 = 4 m/s (0.5) vx1 = 2 m/s x = (2 m/s) (1.07s) x = 2.14 m (distance travelled)

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

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