XOLANI ERIC MAKONDO

• VERTICAL PROJECTILE MOTION

Skydiver diving off cliff

What happens during the motion of a ball that is thrown upwards into the air?

Describe the motion - • The object starts with maximum velocity as

it leaves the throwers hand• The object slows down as it rises in the air• The object momentarily stops at the top• The object speeds up as it descends• The final velocity of the object when it

again reaches the throwers hand is the same as when it left the throwers hand

• At all times the object accelerates downwards due to the force of gravity.

Time for the downward journey = t

Time for the upward journey = t

a = g = 10 m/s2 down

a = g = 10 m/s2 down

vf = 0

vi = maximum value up

vf = maximum value down

vi = 0

ACCELERATION

• At any point during the journey the acceleration of the object is equal to the gravitational acceleration, g.

• g = 10m/s2 down towards the earth.• g is independent of the mass of the object.• g is dependent upon the

distance from the centre of the earth

Use equations of motion -

vf = vi + gt

Δx = vit + ½ gt2

vf2 = vi

2 + 2gs

Δx = (vf + vi) . t 2

t

s/m

GRAPHS of MOTION for PROJECTILE MOTION

tg/m/s2

t

s/m

tv/m/s

Gradient of the graph = - 10m/s2

a = g= -10m/s2 down

Disp./timeVelocity/time

Acceleration/timeDistance/time

FREE FALL and TERMINAL VELOCITY

• Objects only accelerate downwards at 10 m/s2 (or 9,8m/s2) in a vacuum near earths surface.

• In air, air resistance increases and decreases acceleration to values less than 10 m/s2.

• Smooth objects experience less air resistance and a = g initially for all objects.

• If air resistance is large and increasing, acceleration decreases to zero and the object falls at constant velocity, called terminal velocity

When the parachutist jumps, her acceleration is 10 m/s2 downwards. The only force acting on her is the force of the earth.

Fres is downwardsAs her velocity increases, so does the force of air resistance opposing the downward force of gravity.

Fres is still downwards, but smaller.

Air resistance is due to collisions with the particles of air.

The greater the velocity of the parachutist, the greater the number of collisions and the greater the air resistance.

Fres decreases until it equals zero and a = 0.

The parachutist now falls with a constant velocity – called terminal velocity

What are you expected to do?• Apply equations of motion to vertical

motion.• Use graphs of motion to describe vertical

motion.• Explain why some objects reach terminal

velocity when falling in the gravitational field.

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

• Choose a direction as positive.• Decide on the time interval that is relevant

to the question.• Write down known values - vi, vf, a, ∆x and

t.• If an object is released or dropped by a

person that is moving up or down at a certain velocity, the initial velocity of the object equals the velocity of that person.

Apply your knowledge!!!

A bullet is fired vertically upwards at 200m/s. Ignore the effects of air resistance, and calculate:

a. the maximum height reached.b. the time taken for the bullet to be at a

height of 1500m on its way down.c. at what height it will be moving at 100m/s

upwards.

This is how your answer should look:

a. Let up be positive for all answers

vi = +200 m/svf = 0 m/sg = -10 m/s2

∆x = ?vf

2 = vi2 + 2g∆x

0 = (200)2 + 2(-10) ∆x ∆x = + 2000m or 2000m up

b. Consider the time period from when the bullet was fired until it is 500m above the starting position.

vi = +200m/s g = -10m/s2

∆x = +500m t = ?

∆x = vit + ½ gt2

+500 = (+200)t + ½ (-10)t2

t = 10s or 30s

30s is when the bullet is on the way down.

c. Consider the time interval from when the bullet is fired until it has a velocity of 100m/s upwards

vi = +200m/svf = +100m/sg = -10m/s2

vf 2 = vi2 + 2g ∆x

(+100)2 = (+200)2 + 2(-10)s

s = 1500m

The acceleration due to gravity is 9.8 m.s-2. It can differ from point to point on the earths surface – depending on the distance from the centre of the earth. All objects fall at this rate – irrespective of their mass.However, we usually take it (g) as 10 m.s-2.Discuss how the acceleration due to gravity could be determined by using a ticker tape and ticker timer.

Position of object

Time/s Displacement/m Velocity/m.s-1 Acc. Due to gravity/m.s-2

0 0 0 10

1 5 10 10

2 20 20 10

3 45 30 10

4 80 40 10

Fal li n

g ob

ject

Calculate the acceleration due to gravity from these values

How can the value of g be determined by using the set up above and the equation ∆x = vit + 1/2 g t2 ?

Pendulum method

Since the pendulum starts from rest, vi = 0 m.s-1∆x = 1/2 g

t2 The pendulum falls about 84 cm. in the time the metre rule falls through ¼ of a swing. Take the time for 20 swings of the pendulum and then divide by 80 to find the time for ¼ of a swing.Substitute into equation and solve for g.

In a vacuum all objects, irrespective of mass, shape or size, fall at the same rate of 9.8m.s-2

In reality, ‘g’ varies from point to point on the earth’s surface. This depends upon:

1. The change in radius from point to point.

2. The varying density of the earth’s surface from point to point.

At the poles ‘g’ is greater and less at the equator.