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[SOLVED] newtons 2nd law

dwsmith

Well-known member
Feb 1, 2012
1,673
I haven't taken a physics course in years so I am little lost.

A ball is thrown with initial speed $v_0$ up an inclined plane. The plane is inclined at an angle $\phi$ above the horizontal, and the ball's initial velocity is at an angle $\theta$ above the plane. Choose axes with $x$ measured up the slope, $y$ normal to the slope, and $z$ across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance $R = \frac{2v_0^2\sin\theta\cos(\theta + \phi)}{g\cos^2\phi}$ from its launch point. Show that for given $v_0$ and $\phi$, the maximum possible range up the inclined plane is $R_{\max} = \frac{v_0^2}{g(1 + \sin\phi)}$.

Here is my drawing:

Newton's $2^{\text{nd}}$ is $\vec{F} = m\vec{a}$ where we can write $\vec{F} = F_x\vec{x} + F_y\vec{y}$.
\begin{alignat*}{3}
F_x & = & m\ddot{x}\\
F_y & = & m\ddot{y}
\end{alignat*}

Now what should I do? Also, am I correct so far?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
what we are looking for is $x(t)$ and $y(t)$.

we're not interested in the mass of the ball, just its acceleration, which is independent of its mass (not "really" but we're assuming we are on the earth, and so close to the center of mass of the earth that its gravitational force overwhelmingly predominates).

this is actually a differential equation in disguise: we're given an acceleration $\mathbf{g}$ and we want to find:

$\mathbf{r}(t) = (x(t),y(t))$

given that $\mathbf{\ddot{r}}(t) = \mathbf{g}$

as with any differential equation, to come up with a "specific" solution, we need to specify some initial values. we can set up our coordinate system so that:

$\mathbf{r}(0) = (0,0)$

and we are given that:

$\mathbf{\dot{r}}(0) = (v_0\cos\theta,v_0\sin\theta)$ <--no dependence on $t$, the ball is launched with a constant velocity.

the tricky part is expressing $\mathbf{g}$ in terms of the $x$- and $y$- axes. you may wish to confirm for yourself that:

$\mathbf{g} = (-g\sin\phi,-g\cos\phi)$

(my notation may be a bit off, it's been nearly 30 years).

we can solve this by integrating, obtaining:

$x(t) = (v_0\cos\theta)t - \left(\dfrac{g}{2}\sin\phi\right)t^2$
$y(t) = (v_0\sin\theta)t - \left(\dfrac{g}{2}\cos\phi\right)t^2$

to find where the ball lands, set $y(t) = 0$ and solve for $t$ (discount the solution $t = 0$, the launch point). plug this value into $x(t)$.

to find $R_{\max}$, regard $R$ as a function of $\theta$ and find $R'(\theta)$. set this equal to 0 and solve for $\theta$. then plug that value back into $R$. i could finish this for you, but it seems as if you'd like to do SOME of the work. :p
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It's been many years since I had two semesters of physics, but here is the method I would use (clumsy as it may be). I show it in full as I presume you are to use vector calculus.

I would write (use the 3 given axes):

$\displaystyle \vec{F}=\vec{F_x}+\vec{F_y}+\vec{F_z}$

where:

$\displaystyle \vec{F_x}=m\frac{d^2x}{dt^2}=-mg\sin(\phi)$

$\displaystyle \vec{F_y}=m\frac{d^2y}{dt^2}=-mg\cos(\phi)$

$\displaystyle \vec{F_z}=m\frac{d^2z}{dt^2}=0$

and so:

$\displaystyle \frac{d^2}{dt^2}x(t)=-g\sin(\phi)\,\therefore\,\frac{dx}{dt}=-g\sin(\phi)t+v_0\cos(\theta)\,\therefore\,x(t)=-\frac{1}{2}g\sin(\phi)t^2+v_0\cos(\theta)t$

$\displaystyle \frac{d^2}{dt^2}y(t)=-g\cos(\phi)\,\therefore\,\frac{dx}{dt}=-g\cos(\phi)t+v_0\sin(\theta)\,\therefore\,y(t)=-\frac{1}{2}g\cos(\phi)t^2+v_0\sin(\theta)t$

$\displaystyle z(t)=0$

When the ball lands, then $\displaystyle y(t)=0$ where $\displaystyle t\ne0$ which implies:

$\displaystyle -\frac{1}{2}g\cos(\phi)t+v_0\sin(\theta)=0$ and so:

$\displaystyle t=\frac{2v_0\sin(\theta)}{g\cos(\phi)}$

The distance the balls lands from its launch point is then:

$\displaystyle x\left(\frac{2v_0\sin(\theta)}{g\cos(\phi)} \right)=-\frac{1}{2}g\sin(\phi)\left(\frac{2v_0\sin(\theta)}{g\cos(\phi)} \right)^2+v_0\cos(\theta)\left(\frac{2v_0\sin( \theta)}{g\cos(\phi)} \right)=$

$\displaystyle \frac{2v_0^2\sin(\theta)(\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi))}{g\cos^2(\phi)}=\frac{2v_0^2\sin(\theta)\cos(\theta+\phi)}{g\cos^2(\phi)}$

Now, to find the maximum range, the only variable we have to play with is $\displaystyle \theta$, and so we want:

$\displaystyle R(\theta)=\frac{2v_0^2\sin(\theta)\cos(\theta+\phi)}{g\cos^2(\phi)}$

$\displaystyle \frac{dR}{d\theta}= \frac{2v_0^2}{g\cos^2(\phi)}\left(-\sin(\theta)\sin(\theta+\phi)+\cos(\theta)\cos( \theta+\phi) \right)$

Equating this to zero implies:

$\displaystyle \sin(\theta)\sin(\theta+\phi)=\cos(\theta)\cos( \theta+\phi)$

$\displaystyle \tan(\theta)=\cot(\theta+\phi)$

$\displaystyle \theta=\frac{\pi}{2}-(\theta+\phi)$

$\displaystyle \theta=\frac{\pi-2\phi}{4}$

and so we have:

$\displaystyle R_{\text{max}}=R\left(\frac{\pi-2\phi}{4} \right)=\frac{2v_0^2\sin\left(\frac{\pi-2\phi}{4} \right)\cos\left(\frac{\pi-2\phi}{4}+\phi \right)}{g\cos^2(\phi)}$

$\displaystyle \frac{2v_0^2\sin\left(\frac{\pi-2\phi}{4} \right)\cos\left(\frac{\pi+2\phi}{4} \right)}{g\cos^2(\phi)}$

Using the product to sum identity $\displaystyle \sin(A)\cos(B)=\frac{\sin(A+B)+\sin(A-B)}{2}$ we have:

$\displaystyle \frac{2v_0^2\left(\frac{\sin\left(\frac{\pi}{2} \right)-sin(\phi)}{2} \right)}{g\cos^2(\phi)}=\frac{v_0^2(1-\sin(\phi)}{g(1-\sin^2(\phi)}=\frac{v_0^2}{g(1+\sin(\phi))}$

edit: I guess with all my previews I didn't notice that a very similar method was already posted. (Tmi)
 
Last edited:

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
yes, that was the answer i was hinting at. z is irrelevant, so i omitted it from the calculations (all the action takes place in the xy-plane).

rather than use tangents and cotangents to solve $R'(\theta) = 0$

i find it is easier to observe that:

$\cos\theta\cos(\theta+\phi) - \sin\theta\sin(\theta+\phi) = \cos(2\theta + \phi)$

and that $\cos$ is injective on the range $[-\pi,\pi]$ (there are physical reasons for limiting ourselves to this range).

but the rest of what you posted is almost stroke-for-stroke what i have on my "scratch sheet".

interestingly enough, $R$ is undefined if $\phi = \dfrac{\pi}{2}$ leading one to suspect if you throw a ball away from a (vertical) wall at some angle $\theta$ it won't ever land on the wall at all (given, of course, a large enough room, and a "not-very-bouncy" ball)! who knew?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
As a student, I was curious after having found that a launch angle of 45° above the horizontal maximizes the range of a projectile (constant gravity, no drag), what this optimal angle would be if the terrain was not horizontal, but sloped away at a constant angle. Here is how I solved it:

Suppose you are on flat ground and you launch a projectile from the ground at an angle of inclination above the horizontal $\displaystyle θ$ with an initial speed $\displaystyle v_0$. Ignoring drag (air resistance), what launch angle will maximize the projectile's range?

Let us orient an xy-coordinate plane coinciding with the projectile's motion, with the origin at the projectile's initial position. Let g represent the acceleration due to gravity. Resolving the projectile's acceleration into its vertical and horizontal components, we have:

$\displaystyle \frac{d^2}{dt^2}y(t)=-g,\,\frac{d}{dt}y(0)=v_0\sin(\theta),\,y(0)=0$

Solving the initial value problem gives:

$\displaystyle \frac{dy}{dt}=-gt+v_0\sin(\theta)$

$\displaystyle y(t)=-\frac{1}{2}gt^2+v_0\sin(\theta)t$

$\displaystyle \frac{d^2}{dt^2}x(t)=0,\,\frac{d}{dt}x(0)=v_0\cos(\theta),\,x(0)=0$

Solving the initial value problem gives:

$\displaystyle \frac{dx}{dt}=v_0\cos(\theta)$

$\displaystyle x(t)=v_0\cos(\theta)t$

Thus, we have the parametric equations describing the projectile's motion:

(1) $\displaystyle y(t)=-\frac{1}{2}gt^2+v_0\sin(\theta)t$

(2) $\displaystyle x(t)=v_0\cos(\theta)t$

To get y as a function of x, we need to eliminate the parameter t. Solving (2) for t, we get:

$\displaystyle t=\frac{x}{v_0\cos(\theta)}$

Substituting for t into (1) gives:

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

To find the range of the projectile, we set $\displaystyle y=0$, and take the non-zero root for x:

$\displaystyle x=\frac{v_0^22\sin(\theta)\cos(\theta)}{g}$

Using the double-angle identity for sine, we have:

$\displaystyle x=\frac{v_0^2\sin(2\theta)}{g}$

We know that the sine function has its maximum value when its angle is $\displaystyle \frac{\pi}{2}$, thus:

$\displaystyle 2\theta=\frac{\pi}{2}\,\therefore\,\theta=\frac{ \pi}{4}$

So, we see a launch angle of 45° maximizes the range of the projectile.

What if, instead of being on flat terrain, we launch the projectile on a hill having constant slope? What angle do we use now? Let $\displaystyle -\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ be the angle measured above the horizontal of the angle of the hillside. In the xy-plane we oriented before, the surface of the hill may be represented by the line:

$\displaystyle y=\tan(\alpha)x$

Recall we have:

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

This gives:

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

Since we disregard the root x = 0, we have:

$\displaystyle x=\frac{v_0^2}{g}\left(\sin(2\theta)-2\tan(\alpha)\cos^2(\theta) \right)$

Differentiating x with respect to θ, we have:

$\displaystyle \frac{dx}{d\theta}=\frac{2v_0^2}{g}\left(\cos(2 \theta)+\tan(\alpha)\sin(2\theta) \right)$

Equating to zero results in:

$\displaystyle \cos(2\theta)+\tan(\alpha)\sin(2\theta)=0$

$\displaystyle \tan(2\theta)=-\cot(\alpha)$

Using the identity $\displaystyle \tan\left(x+\frac{\pi}{2} \right)=-\cot(x)$ gives

$\displaystyle \cot\left(\theta-\frac{\pi}{2} \right)=\cot(\alpha)$

Equating angles gives:

$\displaystyle 2\theta-\frac{\pi}{2}=\alpha$

Solving for θ, we have:

$\displaystyle \theta=\frac{\pi}{4}+\frac{\alpha}{2}$

So we see we add half the angle of inclination of the hill to 45° to maximize the range of the projectile.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
so, if you're shooting down-hill...."aim a little low"...and if you're on a wall, just drop the darn thing.