Revision:Projectile Motion (M2)

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Projectile Motion

This follows on from M1 Kinematics. Note this is in M1 for the AQA syllabus MM1A and MM1B!

Equations of motion

Consider a particle projected at a speed u at an angle above the horizontal. The only force assumed to be acting upon it is the force due to gravity.

Thus we can resolve the velocity in two components and apply the equation in each case.

The horizontal component of the velocity is . There is no force acting on the particle horizontally, so , where x is the displacement horizontally.

The vertical component of the velocity is . However there is also an acceleration vertically downwards of . Therefore, , where y is the displacement vertically.

Finding the range

The range is the total distance travelled horizontally. This can be found by calculating the values of x for which .

So putting equal to zero, we obtain

So either or

Obviously t = 0 is at the start of the motion. So input the value of t into the equation for x to obtain

Using the double angle formula , this reduces to

And this is the range.

Finding the time of flight

The time of flight can be found by calculating the value of t when x = range.

So using and

we get

The and a u will cancel, leaving

time of flight =

The equation of motion of a projectile

The equations and can be combined by eliminating t to give one equation in x, y and .

Expressing in terms of t, we get

Substituting in  :

Which tidies up to give

Then

Using the identity

the equation of motion of a projectile is