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Revision:Projectile Motion (M2)

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics Revision Notes > Projectile Motion

Contents

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 \theta 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 s = ut + \frac{1}{2} at^2 in each case.

The horizontal component of the velocity is u cos\theta. There is no force acting on the particle horizontally, so x = ut cos\theta, where x is the displacement horizontally.

The vertical component of the velocity is u sin\theta. However there is also an acceleration vertically downwards of g ms^{-1}. Therefore, y = ut sin\theta - \frac{1}{2} gt^2, 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 y = 0.

So putting y = ut sin\theta - \frac{1}{2} gt^2 equal to zero, we obtain

ut sin\theta - \frac{1}{2} gt^2 = 0

So either t = 0 or t = \frac{2u sin\theta}{g}

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

x = \frac{2u^2 sin\theta cos \theta}{g}

Using the double angle formula sin2\theta = 2sin\theta cos\theta, this reduces to

x = \frac{u^2 sin2\theta}{g}

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 x = \frac{u^2 sin2\theta}{g} and t = \frac{2u sin\theta}{g}

we get \frac{2u^2 sin\theta cos\theta}{g} = ut cos\theta

The cos \theta and a u will cancel, leaving

time of flight = \frac{2u sin\theta}{g}


The equation of motion of a projectile

The equations y = ut sin\theta - \frac{1}{2} gt^2 and x = ut cos\theta can be combined by eliminating t to give one equation in x, y and \theta.

Expressing x = ut cos\theta in terms of t, we get t = \frac{x}{u cos\theta}

Substituting in y = ut sin\theta - \frac{1}{2} gt^2 :

y = u sin\theta \times \frac{x}{u cos\theta} - \frac{1}{2}g {(\frac{x}{u cos\theta})}^2

Which tidies up to give y = x tan\theta - \frac{gx^2}{2u^2 cos^2\theta}

Then y = x tan\theta - \frac{gx^2 sec^2\theta}{2u^2}

Using the identity 1 + tan^2\theta = sec^2\theta

the equation of motion of a projectile is y = x tan\theta - \frac{gx^2 (1 + tan^2\theta)}{2u^2}

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