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CfE Higher Physics

CfE Higher Physics. Unit 1 Our Dynamic Universe Projectiles. Learning Intentions. I can carry out calculations to find the horizontal and vertical components of vectors using the relationships: V H = VcosӨ Vv = VsinӨ. W = mg. Projectiles.

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CfE Higher Physics

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  1. CfE Higher Physics Unit 1 Our Dynamic Universe Projectiles

  2. Learning Intentions • I can carry out calculations to find the horizontal and vertical components of vectors using the relationships: VH = VcosӨVv = VsinӨ

  3. W = mg Projectiles A projectile is any object that has been thrown/fired and is under the influence of a gravitational field. The only force acting on a projectile is its weight acting vertically downwards. A projectile therefore accelerates in the downward direction. Projectiles follow a curved path.

  4. Solving Numerical Problems The horizontal and vertical motions of a projectile should be considered separately. • Horizontally - constant velocity (no forces) (v=d/t only) • Vertically - constant downward acceleration (due to weight) (SUVAT only) For the horizontal motion we can use the displacement, velocity, time formula (velocity is constant).

  5. Solving Numerical Problems • For the vertical motion we must use equations of motion (constant acceleration). • The horizontal and vertical motion are linked by time.

  6. Hints and Tips: • Acceleration due to gravity is always negative. • If the object travels down, displacement is negative. • Time taken to travel horizontally is the same vertically. • The time to reach its max height is half the full time of flight. • Split into horizontal and vertical motions. • When you use VcosӨ or VsinӨ you are calculating the initial velocity horizontally or vertically! • The vertical velocity at the max height is 0 ms-1

  7. Horizontal Vertical d = v x t E.O.M. s = ? u = v sin θ v = ? a = - 9.8 m s-2 t = ? t d = ? v =v cos θ t = ? Projectiles vv = v sin θ v OR θ vH = v cos θ

  8. θ θ d1 d2 More Rules on Projectiles… t= ½ total time v= 0 m/s • At maximum height, vertical speed is zero (v=0) • At maximum height, the horizontal distance is half the final value • At maximum height, the time taken is half the final value

  9. 42 ms-1 35° Worked Example 1 A golf ball leaves the tee with a velocity of 42 ms-1 at an angle of 35° to the ground. How far does the ball travel before it hits the ground? Initial horizontal velocity = 42cos35° uh = 34.4 ms-1 Initial vertical velocity = 42sin35° uv = 24.1 ms-1

  10. returns to ground leaves tee Worked Example 1(continued) Vertical motion: sv = uvt + ½at2 0 m sv = 0 = 24.1t+ ½(-9.8)t2 24.1 ms-1 uv = 0 = 24.1t- 4.9t2 vv = -9.8 ms-2 a = 0 = t(24.1 – 4.9t) t = ? t = 0 s or 4.92 s upwards is positive

  11. Worked Example 1(continued) Horizontal motion: sh = vht = 34.4 x 4.92 = 169 m

  12. Worked Example 2 A car travelling at 16 ms-1 drives horizontally off the edge of a 62 m cliff. With what velocity does the car hit the water? Vertical motion: vv2 = uv2 + 2asv -62 m sv = 0 ms-1 uv = vv2 = 02 + 2(-9.8)(-62) ? vv = vv2 = 1215 9.8 ms-2 a = t = vv = 34.9 ms-1

  13. 16 ms-1 34.9 ms-1 Worked Example 2(continued) Horizontal motion: vh = 16 ms-1 Combining the two: magnitude = √(34.92+ 162) = 38.4 ms-1 direction, θ = tan-1(34.9/16) = 65.4° to horizontal θ

  14. Exam Style Questions – stick into notes

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