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Volume 4: Mechanics 1 Vertical Motion under Gravity.

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Presentation on theme: "Volume 4: Mechanics 1 Vertical Motion under Gravity."— Presentation transcript:

1 Volume 4: Mechanics 1 Vertical Motion under Gravity

2 If we drop a stone, it falls towards the Earth. The gravitational force produces an acceleration which at the Earth’s surface is approximately If we model the stone as a particle and ignore air resistance, the only force acting on the stone is Weight is the force due to gravity. acceleration, a = g = 9·8 m s -2 To solve problems involving vertical motion under gravity, we can use the equations of motion for constant acceleration. 9·8 m s -2. its weight.

3 There are 2 types of problems 1)Object initially thrown up – hence it initially slows down acc = -9.8ms -2. Make UP +ve 2) Object initially thrown down or dropped – hence it speeds up acc = +9.8ms -2 Make DOWN +ve What happens initially is crucial – not whether it comes down later.

4 e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground. Solution: s, u, v, a, t

5 Solution: e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground.

6 s, u, v, a, t Solution: e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground. a = g = +9·8 m s -2 as initially dropped. Make DOWN +ve

7 s, u, v, a, t Solution: ? e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground. a = g = 9·8 m s -2

8 s, u, v, a, t Solution: ? e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground. a = g = 9·8 m s -2

9 s, u, v, a, t Solution: ? v  u  a t s  ut + at s  (u + v)t 2 1 v 2  u 2  2as a = g = 9·8 m s -2 e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground.

10 s, u, v, a, t Solution: We need the equation without t. ? v 2  u 2  2as v 2  0  2(9·8)(2)  v 2  39·2   v  6·26 ( 3 s.f. ) The velocity is 6·26 m s -1. e.g.1A stone falls from rest from a point 2 m above the ground. Modelling the stone as a particle and ignoring air resistance, find its velocity as it hits the ground. a = g = 9·8 m s -2 v  u  a t s  ut + at s  (u + v)t 2 1 v 2  u 2  2as

11 e.g.2I throw a ball vertically upwards so that as it leaves my hand it has a velocity of 14·7 m s -1. Modelling the ball as a particle and ignoring air resistance, find (a) the time taken to reach the maximum height, (b) the velocity and displacement after 2 s. Solution: s, u, v, a, t (a) As the ball will travel up and down, we must be careful with signs. Always draw a diagram. u  14·7 a = g = -9·8 m s -2 as initially thrown up. Make UP +ve

12 Modelling the ball as a particle and ignoring air resistance, find Solution: s, u, v, a, t At the maximum height, the ball is neither going up nor down, so the velocity is zero. (a) v  0 u  14·7 (a) the time taken to reach the maximum height, (b) the velocity and displacement after 2 s. e.g.2I throw a ball vertically upwards so that as it leaves my hand it has a velocity of 14·7 m s -1.

13 Modelling the ball as a particle and ignoring air resistance, find Solution: s, u, v, a, t The ball is thrown up. What is the acceleration ? (a) Ans: The ball is slowing down, so the acceleration is negative. Make UP +ve a  9·8 v  0 u  14·7 (a) the time taken to reach the maximum height, ? (b) the velocity and displacement after 2 s. e.g.2I throw a ball vertically upwards so that as it leaves my hand it has a velocity of 14·7 m s -1.

14 s, u, v, a, t ? a  9·8 v  0 0  14·7  (  9·8)t  9·8t  14·7   t  14·7 9·8 u  14·7  t  1·5 s v  u  a t

15 Graph of s=ut +  at 2 Ex s=20t -  at 2

16 s, u, v, a, t ? a  9·8 u  14·7 Since the ball has taken 1·5 s to reach its greatest height, we know that it will be (b) Find the velocity and displacement after 2 s. t  2 ( The up and down motion is in one straight line but it’s clearer if we draw the lines side by side. ) We do not need to change the sign of a. on its way down after 2 s.

17 s, u, v, a, t ? a  9·8 u  14·7 v  u  a t t  2 v v  v   4·9 m s -1  After 2 s the velocity is  4·9 m s -1. ( The ball is coming down. ) (b) Find the velocity and displacement after 2 s. s  ut + at s  14·7(2)  (  9·8)(2) 2  ·7  (  9·8) 2 s  9·8 m  ( After 2 s the ball is 9·8 m above the point where it left my hand. )

18 e.g.2A stone is thrown upwards with a velocity of 5 m s -1 at the edge of a cliff and falls into the sea 60 m vertically below the starting point. Modelling the ball as a particle and ignoring air resistance, find the length of time it is in the air. Solution: s, u, v, a, t u  5 a  10 In this question, assume the value of g is 10 m s -2

19 e.g.2A stone is thrown upwards with a velocity of 5 m s -1 at the edge of a cliff and falls into the sea 60 m vertically below the starting point. Modelling the ball as a particle and ignoring air resistance, find the length of time it is in the air. Solution: s, u, v, a, t u  5 s   60 s  0 ? a  10 In this question, assume the value of g is 10 m s -2 If we assume that the displacement, s, is zero when the stone is thrown, s will be negative below this level. Make UP +ve

20 Solution: s, u, v, a, t u  5 s   60 s  0 ? a  10 s  ut + at  60   2 1 t   3 or  4   12  t  t 2 Divide by 5 : (t  3)(t  4)  0  t 2  t  12  0  This is a quadratic equation so we must either factorise or use the formula. We need zero on one side. 5t  (  10) t 2t 2 What meaning we can give to t =  3 Ans:A negative value represents time before the stone was thrown. The stone takes 4 s to reach the sea.

21 e.g.4.A stone is dropped from rest into a well and falls into the water 10 m below. Ignoring all forces except the force due to gravity, find the time taken until it hits the water. Solution: s, u, v, a, t ? s  ut + at u  0 a  9·8 In this question, all the motion is downwards so, Make DOWN +ve s    0t  9·8 t 2t  10  4·9t 2   t  1·43 s  t ·9 ( The negative root is not possible. )

22 1.A ball is dropped from rest at the top of a tower and hits the ground 2 s later. Modelling the ball as a particle and ignoring air resistance, find the height of the tower. EXERCISE Solution: s, u, v, a, t ? s  ut + at u  0 a  9·8 s s   0t  9·8 (2)  t  2 s s  19·6 The tower is 19·6 m high.

23 2.A ball is thrown upwards with a velocity of 12 m s -1 and travels in a straight line landing on the ground 1 m below its starting height. Modelling the ball as a particle and ignoring air resistance, find how fast it is moving as it hits the ground. Solution: s, u, v, a, t u  12 a  9·8 s  1s  1 ? v 2  u 2  2as v 2   12 2  2 (  9·8)(  1) v 2  163·6  v  12·8 m s -1 

24 Past M1 question

25 Answers

26 Notes page 27

27 SUMMARY  For vertical motion where the only force is the weight ( the force due to gravity ) we can use the equations of motion for constant acceleration.  The acceleration due to gravity has a magnitude of 9·8 m s -2 close to the surface of the Earth.  We draw a sketch of the motion to avoid sign errors.  For motion that is all downwards, we usually take the positive axis downwards so g   9·8. Otherwise, the positive axis is up and g   9·8  There is no need to separate the parts of the motion where the body is moving up and down.

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29 The summary page follows in a form suitable for photocopying.

30 Summary VERTICAL MOTION UNDER GRAVITY TEACH A LEVEL MATHS – MECHANICS 1  For vertical motion where the only force is the weight ( the force due to gravity ) we can use the equations of motion for constant acceleration.  The acceleration due to gravity has a magnitude of 9·8 m s -2 close to the surface of the Earth.  We draw a sketch of the motion to avoid sign errors.  For motion that is all downwards, we usually take the positive axis downwards so g   9·8. Otherwise, the positive axis is up and g   9·8.  There is no need to separate the parts of the motion where the body is moving up and down.


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