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Find the range of a projectile on an inclined plane

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Consider the following problem:
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A hill makes an angle $\alpha$ with the horizontal.
An arrow is shot from a point on the hill with initial velocity $v_{0}$ and at an angle $\beta>\alpha$ with the horizontal.

(a) Find the time needed for the arrow to land in terms of $\alpha,\beta,v_{0}$ and the gravitational acceleration.

(b) Show that the distance between the origin and the place of landing is given by

I am unable to get the expected answer for part (b). Since part (b) depends on parts (a) I am imagining my reasoning in (a) is faulty somewhere, but where?

Here it is my answer for part (a)

and for part (b) I get

asked Sep 13 in Physics Problems by member192 (270 points)
retagged Sep 13 by sammy gerbil

1 Answer

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Best answer

Use axes $Ox, Oy$ which are respectively parallel and perpendicular to the inclined plane. The components of the launch velocity along these axes are $v_x=v_0 \cos(\beta-\alpha)$ and $v_y=v_0\sin(\beta-\alpha)$, while the components of gravity are $-g_x=-g\sin\alpha$ and $-g_y=-g\cos\alpha$.

The co-ordinates of the arrow at time $t$ after launch are $$x=v_x t-\frac12 g_x t^2$$ $$y=v_y t-\frac12 g_y t^2$$ When the arrow lands then $y=0$ and $t=T$ the time of flight and $x=R$ the range. So $$v_y=\frac12 g_y T$$ $$T=\frac{2v_y}{g_y}=\frac{2v_0}{g}\frac{\sin(\beta-\alpha)}{\cos\alpha}$$ and $$R=v_xT-\frac12 g_xT^2=\frac{2v_0^2}{g}\frac{\cos(\beta-\alpha)\sin(\beta-\alpha)}{\cos\alpha}-\frac{2v_0^2}{g}\frac{\sin\alpha \sin^2(\beta-\alpha)}{\cos^2\alpha}$$ $$=\frac{2v_0^2\sin(\beta-\alpha)}{g\cos^2\alpha}[\cos(\beta-\alpha)\cos\alpha-\sin(\beta-\alpha)\sin\alpha]$$ $$=\frac{2v_0^2}{g}\frac{\sin(\beta-\alpha)\cos\beta}{\cos^2\alpha} $$

To find the angle of launch for maximum range note that $$2\cos A\sin B = \sin (A+B) - \sin (A-B)$$ therefore $$2\cos\beta\sin(\beta-\alpha)=\sin(2\beta-\alpha)-\sin\alpha$$ The 2nd term is fixed because $\alpha$ is fixed, but $\beta$ can vary. The maximum value this expression can have occurs when the 1st term is $+1$, ie when $$2\beta-\alpha=\frac12\pi$$ $$2(\beta-\alpha)= \frac12 \pi-\alpha$$ That is, the maximum range is achieved when the direction of launch bisects the angle between the inclined plane and the gravity vertical. The maximum range is $$R_{max}=\frac{v_0^2}{g}\frac{(1-\sin\alpha)}{\cos^2\alpha}=\frac{v_0^2}{g}\frac{(1-\sin\alpha)}{(1-\sin\alpha)(1+\sin\alpha)}=\frac{v_0^2}{g(1+\sin\alpha)}$$

Source : Letter to the Editor of American Journal of Physics by A P French, 1984

answered Sep 13 by sammy gerbil (26,096 points)
selected Sep 14 by member192
Thank you for the detailed answer and the link to the paper.
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