Projectiles have components of vertical and horizontal motion which must be analysed individually, as they have no effect on each other. The horizontal velocity of a projectile is constant, whilst the vertical will change as the object accelerates due to gravity.
Equations
DESCRIPTION
EQUATION
Velocity
v=vx2+vy2
Angle from horizontal
θ=tan−1(vyvx)
Constants
CONSTANT
SYMBOL
VALUE
accelerationduetogravity
g
-9.81ms−2
Variables
QUANTITY NAME
SYMBOL
DERIVED UNIT
SI UNIT
xcomponentofvelocity
vx
ms−1
ms−1
ycomponentofvelocity
vy
ms−1
ms−1
anglefromthehorizontal
θ
°
rad
displacement
s
m
m
timetaken
t
s
s
Projectiles
A projectile is an object that is thrown at an angle to the horizontal, and therefore has motion in both the horizontal and vertical plane. The horizontal and vertical motions are independent of each other, so they must be analysed on their own.
However, time is interchangeable between the directions of motion.
In projectile motion, air resistance is assumed to be negligible:
the vertical velocity changes due to the acceleration due to gravity, g.
the vertical components can be calculated using equations of motion using a=g.
the horizontal velocity is constant.
1.
Vertical component
for velocity will
increase with
time
2.
Horizontal component
for velocity will
remain constant
3.
Projectile path
4.
Horizontal displacement
5.
Vertical displacement
Horizontal velocity
The horizontal velocity remains constant as there are no forces acting in the horizontal direction. The horizontal velocity can therefore be analysed using the equation:
vx=tsx
The component of initial horizontal velocity is:
ux=ucos(θ)
Vertical velocity
The vertical velocity changes due to the acceleration of free fall g. The suvat equations of motion can be used to analyse the vertical motion. The initial velocity is:
uy=usin(θ)
Sometimes projectile problems require the journey to be split into two, the first half from launch to max height and then the second half from max height to ground.
Note: The direction of motion must be defined for the whole question.
Resolving velocities
The path of a projectile is commonly curved, as the vertical velocity increases whilst the horizontal velocity remains constant. The magnitude of the velocity can be calculated by using the horizontal and vertical components of velocity:
v=vx2+vy2
The angle made by the velocity to the horizontal is given by:
θ=tan−1(vyvx)
Example
An object is launched at a velocity of 40ms−1 at an angle of 50° above the horizontal. Calculate the maximum height reached.
Firstly, considering the vertical components, state the variables in the question: