Q2.3 The trajectory equation (S)

The equation describing the trajectory of a particle launched at speed v and angle θ to the horizontal is (Equation 1.10)

\[ y= x \tan \theta -\frac{gx^2}{2v_0^2 \cos ^2 \theta} \]

Using this equation establish

Sketch the trajectory.

One can acccount for air resistance (‘drag’) by allowing for an additional contribution to the acceleration, which acts back along the path. Show qualitatively (with a further sketch) what the effect will be on the shape of the trajectory.

Hide

Hint

What condition is satisfied by y at the end of the flight?

What is the derivative of y with respect to x at the highest point?

If there is air resistance what will the horizontal component of velocity be in the later stages of the path? (Suppose the particle is in the air for a ‘long’ time.)

 

Solution

Reveal
Hide

Solution

  • [(a)] The condition locating the end of the flight is y=0. Inserting this in the trajectory equation gives
    \[ 0= x \tan \theta -\frac{gx^2}{2 v_0^2 \cos ^2 \theta} \]
    which has the solutions x=0 (identifying the launch point) and
    \[ x= \frac{2v_0^2\cos ^2 \theta \tan \theta}{g} = \frac{2v_0^2\cos \theta \sin \theta}{g} =R \]
    recovering Equation 1.9
  • [(b)] At the highest point (by definition) y has a maximum so that $\frac{dy}{dx}=0$. Inserting this in the trajectory equation gives
    \[ 0= \tan \theta -\frac{gx}{v_0^2 \cos ^2 \theta} \]
    implying
    \[ x=\frac{v_0^2\cos \theta \sin \theta}{g}=\frac{R}{2} \]

The trajectory equation describes a parabola, which is symmetric about the line x=R/2.

If we accommodate air resistance as suggested we recognise that the x component of velocity will no longer be constant, but will reduce throughout the flight. Towards the end of the flight, in particular, it will be small, even zero. At this stage the projectile will be falling vertically.

Graphic - No title - trajeqn

As a result of the additional downward acceleration acting in the early stages of the flight (when the projectile is moving upwards) the projectile will no longer reach the same height. The trajectory should then have the qualitative form shown.

Aside: Once you’ve got some more mathematics under your belt, you will be able to solve the equations that describe the motion, with air-resistance, quantitatively.