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Chapter 1

This document discusses equations for projectile motion. It provides the equations to treat horizontal and vertical motion separately under constant acceleration due to gravity. An example is given of a ball projected at an angle to illustrate the horizontal and vertical components of motion.

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61 views39 pages

Chapter 1

This document discusses equations for projectile motion. It provides the equations to treat horizontal and vertical motion separately under constant acceleration due to gravity. An example is given of a ball projected at an angle to illustrate the horizontal and vertical components of motion.

Uploaded by

Wuileap
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Equations for projectile motion

A projectile moves in two dimensions under the action of only one force, the
force of gravity, which is constant and acts vertically downwards. This means
that the acceleration of the projectile is gms-2 vertically downwards and
there is no horizontal acceleration.You can treat the horizontal and vertical
motions separately, using the equations for constant acceleration.

Horizontal distance travelled is small Vertical distance travelled is small


enough to assume that gravity is always enough to assume that gravity is
l&I
..J
in the same direction. constant.

ti.
a:
l&I :

a:: · Note
A.
◄ The value of g varies around the world, from 9.766 in Kuala Lumpur to 9.825
IL in Oslo. A value of 10 is used in this book.
0
z
0
To illustrate the ideas involved, think of a ball being pro jected with a speed
Note of20m s-1 at 60° to the ground, as illustrated in Figure 1.1.This could be a
You have already first model for a football, a chip shot from the rough at golf or a lofted shot
worked with at cricket.
vectors in Pure
Mathematics 3. acceleration
In this chapter
they are used to height
make it easier y(m)
15
to distinguish
between motion
in the horizontal 10
and vertical
directions. 20sin 60"
5
Although the
Cambridge
International 0 ...___........ 5 __2._
1 o__....
5 __.... 30____3__
__5__.... 5 - x---
1__ 0__2 ( m) 20 cos 60"
syllabus does not horizontal distance ...._________,
require students
to use vector A Figure 1.1
methods, vectors
Using axes as shown, the components are:
can provide a
useful way to Horizontal Vertical This is negative because
simplify and the positive y-axis is
solve mechanics Initial position 0 0 upwards. As a vector
problems. Acceleration aX = 0 a =-10
y
a=(-1�)
Initial velocity ux =20 cos60° u =20sin60°
= 10 = 17.3... 4
As a vector
20 cos 60" "17.3... · means 17.3 and subsequent figures. Keep this number in your calculator
u= ( ) for use in future working. You should not round values that will be used in
20 sin 60°
later calculations. Convention for this course is to give final answers to three
significant figures (or one decimal place for angles in degrees). unless a question
asks for something different.

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