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Linear Accelerated Motion Part 2 For the Higher Level Leaving Cert Course ©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork.

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Presentation on theme: "Linear Accelerated Motion Part 2 For the Higher Level Leaving Cert Course ©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork."— Presentation transcript:

1 Linear Accelerated Motion Part 2 For the Higher Level Leaving Cert Course ©Edward Williamson, Applied Maths Local Facilitator, Coachford College, Co Cork

2 Section 3 Vertical Motion

3 Gravity Signs are always how you start e.g. if you start throwing upwards, then up is + Acceleration is g and always down. Sign depends on if you say up is + or down is +. Watch for displacement e.g. throwing something off a cliff 30 m above ground, hence displacement is -30

4 Example 3.1 [LC:1988 Q1 (b)] A particle falls freely from rest from a point o, passing three points a, b and c, the distances ab and bc being equal. If the particle takes 3 s to pass from a to b and 2 s from b to c, calculate |ab|

5 A B C u=uu=u + x 2x2x Always go from A as initial velocity remains same A to B u=u, a=g, s=x, t=3 A to C u=u, a=g, s=2x, t=5 Solve to give x =147 m Down is positive hence g is positive here

6 Example 3.2 [LC:2002 Q1(a)] A stone is thrown vertically upwards under gravity with a speed of u m/s from a point 30 m above the horizontal ground. The stone hits the ground 5 s later. i) Find the value of u ii) Find the speed with which it hits the ground.

7 u=+u a=-g s=-30 t=5 Finding u Finding v Displacement is -30 because up is positive

8 Example 3.3 [LC:1990 Q1(a)] A particle is projected vertically upwards with velocity u m/s and is at a height h after t 1 and t 2 seconds respectively. Prove that:

9 h + u=+u a=-g s=+h t=t (t 1 on way up, t 2 on way down) Solving Product of 2 roots Remember α and β

10 Example 3.4 [LC:1992 Q1(a)] A balloon ascends vertically at uniform speed. 7.2 seconds after it leaves the ground, a particle is let fall from the balloon. The particle takes 9 seconds to reach the ground. Calculate the height from which the particle was dropped.

11 u h + Balloon For the Particle u= +u a= -9.8 t= 9 s= -h

12 Example 3.5 [LC:2007 Q1(a)]

13 3 rd second is between t=2 seconds and t=3 seconds Let x = distance travelled in first 3 seconds Let y = distance travelled in first 2 seconds From the Question Equations always give zero to n time

14 Finding Height of the tower

15 Now try some questions by yourself on the attached sheet


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