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Projectile motion

These notes present the solution of the problem #43 from Ch. 3 of Serway&Jewett.

A projectile is launched up an incline (incline angle $\phi$) with an initial speed $v_i$ at an angle $\theta_i$ with respect to the horizontal $(\theta_i > \phi_i)$, (see the setup on the right of Fig. 1).

(a) Show that the projectile travels a distance $d$ up the incline, where

(b) For what value of $\theta_i$ is $d$ a maximum, and what is that maximum value?

Review of the task: (a) to check that the distance along the incline $d=\vert OP\vert$ is given by a certain formula, and (b) to find the value of $\theta$ which, for a fixed $\phi$, gives us the maximum value of the range $d$.

Figure 1: The setup. Some notations have been changed for the sake of clarity.

To solve the problem we have at least two choices for the co-ordinate system. The choice (I) has the $x$ axis along the baseline of the incline, while the second choice (II) has the $x$ axis along the incline itself (see Fig. 1).

We write the kinematic equations the usual way. For the case (I) we have

$$x = (v_0\cos\theta)t,$$ (1)

$$y = (v_0\sin\theta)t -\frac{g}{2}t^2.$$ (2)

It's obvious that the point of impact with the incline (P) is a point on the trajectory $y=Y(x)$, with $Y$ being the function one gets by eliminating $t$ using (1) and (2). By doing so we obtain

$$y = x\tan\theta -\frac{g}{2}\frac{x^2}{v_0^2\cos\theta}$$ (3)

The coordinates of P $(x_P,y_P)=(d\cos\phi,d\sin\phi)$ have to satisfy (3), and hence we get

$$d\sin\theta = d\cos\theta\tan\theta -\frac{g}{2}\frac{d^2\cos^2\theta}{v_0^2\cos^2\theta}.$$ (4)

Equation (4) has two solutions. One of them is $d=0$ as expected. The other one is¹

$$d = - \frac{2v_0^2\cos^2\theta}{g\cos^2\phi}(\sin\phi -\cos... ...ta) = \frac{2v_0^2\cos\theta\sin(\theta-\phi)}{g\cos^2\phi},$$ (5)

which coincides with the expression provided.

For the case (II) the equations of motion read

$$x = [v_0\cos(\theta-\phi)]t-\frac{g\sin\phi}{2}t^2,$$ (6)

$$y = [v_0\sin(\theta-\phi)]t -\frac{g\cos\phi}{2}t^2.$$ (7)

Apparently we got a complication, since both $x$ and $y$ are now quadratic in $t$. Fortunately, in the case (II) the condition which gives us the distance $d$ is particularly simple, and removes the above mentioned obstacle. Namely, one observes that for the choice (II) we have $y=0$ only at points O (where $x=0$) and P (where $x=d$). Equation (7) then gives us

$$0 = t\left(v_0\sin(\theta-\phi) -\frac{g\cos\phi}{2}t\right),$$ (8)

with two solutions for the flight time. One is the trivial $t=0$, the other one, corresponding to the impact time at P is

$$t = \frac{v_0\sin(\theta-\phi)}{g\cos\phi}.$$ (9)

Substituting (9) into (6) (see also footnote 1) we get the distance $d$ along the incline, namely

$$d = \frac{2v_0\sin(\theta-\phi)}{g\cos\phi} \left\{v_0\cos(\... ...ght\}= \frac{2v_0^2\sin(\theta-\phi)\cos\theta}{g\cos^2\phi}.$$ (10)

Moral: both methods worked just fine.

Note: To get $d$ we didn't divide anywhere by expressions which vanish when $\phi= 0$. Hence, we expect that in the limit $\phi\rightarrow 0$ the expression for $d$ recovers a known result for projectile motion (e.g. find the range for given initial $v_0$ and angle $\theta$ with horizontal). In our case, for $\phi= 0$ we get $d=v_0^2\sin2\theta/g$, so the check for a limiting case is passed.

To solve the part (b) we also have at least two choices.

One choice would be to take the derivative of $d$ with respect to $\theta$, while keeping $v_0, \phi, g$ constant, and demand $$\frac{\partial d}{\partial \theta} =0$$. This will give us $\theta$ for which $d$ has an extremum or saddle point. We need then to check that this extremum is indeed a maximum.

The other choice is to rewrite the expression for $d$ in such a way that the answer becomes obvious. We will pursue this path.

Let us note that for given $v_0, \phi, g$ (and kept constants) the only factor in the expression for $d$ that depends on $\theta$ is $\sin(\theta-\phi)\cos\theta$. This can be rewritten as (see footnote 1)

$$\sin(\theta-\phi)\cos\theta = \frac{1}{2} \left\{\sin(\theta... ...ht\}= \frac{1}{2}\left\{\sin(2\theta-\phi)-\sin(\phi)\right\}.$$ (11)

For a fixed $\phi$ the above expression reaches its maximum value for $\sin(2\theta-\phi)=1$, which implies that $2\theta-\phi=\pi/2$. One may easily check that for $\theta, \phi \in (0,\pi/2)$ this is the only solution of $\sin(2\theta-\phi)=1$. Hence, the angle which gives us the maximum $d$ is $$\theta =\frac{\pi}{4}+\frac{\phi}{2}$$. Using (11), the maximum value of $d$ is found to be

$$d = \frac{v_0^2}{g\cos^2\phi}(1-\sin\phi)$$ (12)

Note: In the limit case $\phi\rightarrow 0$, eqn. (12) gives us the well known result $d=v_0^2/g$.

Last revised: September 25, 2003 © 2003, Sorin Codoban