Vertical Projectiles Vertical Projectile Motion Vertical projectile motion

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Vertical Projectiles

Vertical Projectiles

Vertical Projectile Motion � Vertical projectile motion deals with objects that fall straight down,

Vertical Projectile Motion � Vertical projectile motion deals with objects that fall straight down, objects that get thrown straight up and the motion of an object as it goes straight up and then down.

� The acceleration of this object due to gravity is called gravitational acceleration, g,

� The acceleration of this object due to gravity is called gravitational acceleration, g, and is equal to 9, 8 m • s-2 on earth.

An object falling straight down from rest � Consider an object dropped from the

An object falling straight down from rest � Consider an object dropped from the top of a building. � It accelerates at 9, 8 m • s-2 and this is always downwards. � The object has an initial velocity, vi of 0 m • s 1. � The displacement, Δy, of the object is equal to the height from which it falls. � The object reaches maximum velocity, vf, on impact with the ground.

An object projected vertically upwards � Consider an object that is thrown vertically upward.

An object projected vertically upwards � Consider an object that is thrown vertically upward. � As it get higher and higher, it slows down until it stops momentarily at its highest point. � It has an initial velocity, vi, with which it was projected i. e. vi is not 0 m • s-1 � The acceleration is 9, 8 m • s-2 downward.

� At the highest point the object stops and therefore its final velocity vf

� At the highest point the object stops and therefore its final velocity vf = 0 m • s-1, but � the acceleration is still 9, 8 m • s-2. � The object speeds up as it ascends. � The time taken for an object to reach its maximum height is the same as the time it takes to come back.

CLASSWORK 1. One word/term items 1. 1. The force that acts on a body

CLASSWORK 1. One word/term items 1. 1. The force that acts on a body in free fall. 1. 2. Motion of an object near the surface of the earth under the influence of the earth’s gravitational force alone.

2. Multiple Choice 2. 1 Which of the following is a correct statement? Gravitational

2. Multiple Choice 2. 1 Which of the following is a correct statement? Gravitational force is A. applicable only in our solar system B. both an attractive and repulsive force C. directly proportional to the product of the masses involved. D. Directly proportional to both the masses and the radius of the earth.

2. 2 A golf ball is hit vertically upwards. What is the acceleration of

2. 2 A golf ball is hit vertically upwards. What is the acceleration of the ball at the highest point? Ignore the effects of friction. A. 9, 8 m • s-2 upwards B. 9, 8 m • s-2 downwards C. 0 m • s-2 D. 6, 8 m • s-2 downwards

Long question 3. A stone is thrown vertically upwards at an initial velocity of

Long question 3. A stone is thrown vertically upwards at an initial velocity of 24 m • s-1. It reaches its maximum height after 2 s. 3. 1 Describe the motion of the stone in terms of velocity and acceleration. 3. 2 How long from the time it is thrown upward, will it take to come back into the thrower’s hand?

5. 3 What turning 5. 4 What turning is the velocity of the stone

5. 3 What turning 5. 4 What turning is the velocity of the stone at the point? is the acceleration of the stone at the point?

SOLUTIONS 1. 1 Gravitational force/weight 1. 2. Free fall 21. C 2. 2. B

SOLUTIONS 1. 1 Gravitational force/weight 1. 2. Free fall 21. C 2. 2. B

3. 1 Initially the stone has a velocity of 24 m • s 1

3. 1 Initially the stone has a velocity of 24 m • s 1 upwards. Throughout the motion it experiences a constant downward acceleration of 9, 8 m • s-2. its velocity decreases to zero at the turning point. There its direction of motion changes and the velocity increases while moving downwards.

3. 2 4 s 3. 3 0 m • s-1 3. 4 9, 8

3. 2 4 s 3. 3 0 m • s-1 3. 4 9, 8 m • s-2

Equations of motion � Projectiles can have their own motion described by a single

Equations of motion � Projectiles can have their own motion described by a single set of equations for the upward and downward motion.

Acceleration � At any point during the journey the acceleration of the object is

Acceleration � At any point during the journey the acceleration of the object is equal to the gravitational acceleration, g. � g is equal to 9, 8 m • s-2 downwards. � g is independent of the mass of an object.

Use equations of motion � vf = vi + gΔt � Δy = viΔt

Use equations of motion � vf = vi + gΔt � Δy = viΔt + ½ gΔt 2 � vf 2= vi 2 + 2 gΔy � Δy = (vf + vi) • Δt 2

Tips to help you use the equations of motion for the projectile motion �

Tips to help you use the equations of motion for the projectile motion � Choose a direction as positive. � Write down the values of the known vf; vi; g; Δy and Δt � If an object is released or dropped by a person that is moving up or down at a certain velocity the initial velocity of an object equals the velocity of that person.

� Identify which formula to use. � Substitute into the equation. � Interpret the

� Identify which formula to use. � Substitute into the equation. � Interpret the answer.

CLASSWORK 1. A ball is thrown vertically upwards and returns to the thrower’s hand

CLASSWORK 1. A ball is thrown vertically upwards and returns to the thrower’s hand 4 s later. Calculate a) the height reached by the ball b) the velocity with which the ball left the thrower’s hand. c) the velocity with which the ball returned to the thrower’s hand.

2. A grade 12 learner wants to determine the height of the school building.

2. A grade 12 learner wants to determine the height of the school building. He projects a stone vertically upward so that it reaches the top of a building. The stone leaves a learner’s hand at a height of 1, 25 m above the ground, and he catches the stone again at a height of 1, 25 m above the ground. He finds the total time the stone is in the air is 2 s. Calculate the height of the school building if air resistance is ignored.

SOLUTIONS 1. a) Δy = viΔt + ½ gΔt 2 = 0 (2) +

SOLUTIONS 1. a) Δy = viΔt + ½ gΔt 2 = 0 (2) + ½ (9, 8)(2)2 = 19, 6 m

b) vf 2 = vi 2 + 2 gΔy 02 = vi 2 +

b) vf 2 = vi 2 + 2 gΔy 02 = vi 2 + 2(-9, 8)(19, 6) vi =19, 6 m • s-1

2. Take the upward motion as positive vf = vi + gΔt 0 =

2. Take the upward motion as positive vf = vi + gΔt 0 = vi + (-9, 8)(1) vi = 9, 8 m • s-1 upward Δy = viΔt + ½ gΔt 2 = (9, 8)(1) + ½ (-9, 8)(1)2 = 4, 9 m Therefore total height = 6, 15 m

Graphs of projectile motion � Equations of motion are equations that are used to

Graphs of projectile motion � Equations of motion are equations that are used to determine the motion of a body while experiencing a force as a function of time. � These equations apply only to bodies moving in one dimension/straight line with a constant acceleration.

� The body’s motion is considered between two time points: that is, from one

� The body’s motion is considered between two time points: that is, from one initial point and its final point in time. � Motion can be described in different ways: ◦ Words ◦ Diagrams ◦ Graphs

� We use three different graphs ◦ velocity – time graph ◦ acceleration –

� We use three different graphs ◦ velocity – time graph ◦ acceleration – time graph ◦ position – time graph

GRAPHS FOR VERTICAL PROJECTILE MOTION. � Consider a basketball player throwing a ball in

GRAPHS FOR VERTICAL PROJECTILE MOTION. � Consider a basketball player throwing a ball in the air. What goes up must come down. � The ball has downward force acting on it because of gravity. Therefore it will slow down at a rate of 9, 8 m∙s-1 after every second. So we can say that the acceleration is -9, 8 m∙s-2.

� When we tackle problems like this, we use the equations of motion. �

� When we tackle problems like this, we use the equations of motion. � We have to make sure that we get the signs right. We will make upwards positive and downwards negative.

� Consider a ball thrown vertically upwards and it returns to thrower’s hand. �

� Consider a ball thrown vertically upwards and it returns to thrower’s hand. � We can represent these motions graphically. It is important that you understand these graphs.

Worked example � An object is dropped from a hot air balloon which is

Worked example � An object is dropped from a hot air balloon which is ascending at a constant speed of 2 m • s-1. Ignore the effects of air resistance. � a) Calculate how far below the point of release the object will be after 4 s. � b) Draw velocity vs time and acceleration vs time graphs for the motion of the object from the moment it is dropped from the balloon until it hits the ground.

Solution � a) Take up as positive g =- 9, 8 ms-1 vi =

Solution � a) Take up as positive g =- 9, 8 ms-1 vi = 2 ms-1 Δt = 4 s Δy = ? Δy = viΔt + ½ gΔt 2 � = (2)(4) + ½ (-9, 8)(4)2 � = 70, 4 m