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Volume 4: Mechanics 1 Vertical Motion under Gravity

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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.

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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.

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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

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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.

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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

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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

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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

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s, u, v, a, t Solution: ? v u a t s ut + at 2 2 1 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.

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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 2 2 1 s (u + v)t 2 1 v 2 u 2 2as

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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

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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.

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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.

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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

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

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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.

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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 2 2 1 s 14·7(2) ( 9·8)(2) 2 2 1 14·7 ( 9·8) 2 s 9·8 m ( After 2 s the ball is 9·8 m above the point where it left my hand. )

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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

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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

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Solution: s, u, v, a, t u 5 s 60 s 0 ? a 10 s ut + at 2 2 1 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.

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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 2 2 1 u 0 a 9·8 In this question, all the motion is downwards so, Make DOWN +ve s 10 10 0t 9·8 t 2t 2 2 1 10 4·9t 2 t 1·43 s t 2 10 4·9 ( The negative root is not possible. )

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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 2 2 1 u 0 a 9·8 s s 0t 9·8 (2) 2 2 1 t 2 s s 19·6 The tower is 19·6 m high.

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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

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Past M1 question

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Answers

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Notes page 27

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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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The summary page follows in a form suitable for photocopying.

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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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