# Projectile along incline [closed]

A particle is projected with velocity $u$ from the bottom of an inclined plane whose inclination with the horizontal is $\beta$. If afterwards the projectile strikes the inclined plane perpendicular to the surface, find the height of the point struck (distance from the ground). The angle made by the velocity with the incline is $\alpha$.

I got an answer while solving this question that differs from the one in my textbook. I got the correct value for the time period though... This is what I did:

Could someone please try this question? Please help me find the height and also tell me the expression you get for the time taken.

My textbook shows

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## closed as too localized by Waffle's Crazy Peanut, Emilio Pisanty, David Z♦Apr 29 '13 at 16:50

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Hi Rohan, welcome to Physics.SE! Why don't you show us what you've done already? (you can use dollar signs to typeset formulas in LaTeX) It's better to have you find the answer by giving you a hint at where you went wrong than to just give you the answer, which is against the homework policy of this website. –  Wouter Apr 29 '13 at 11:13
I got the time period as 2usin(alpha) / ( gcos(beta) ). Is it possible to show you my working by uploading an image? I used some logical vector analysis that differs from my textbook, which I learnt from my Physics teacher... –  Rohan R Apr 29 '13 at 11:20
Since you don't have enough reputation yet, you can't do that directly. But if you put in the link to the image, I can edit your question and add it as an actual image. –  Wouter Apr 29 '13 at 11:29
@Wouter The link to my working is j.mp/projectincline. –  Rohan R Apr 29 '13 at 11:44
I've edited it in. A rotated image would be easier, though I guess people can also do that themselves... –  Wouter Apr 29 '13 at 11:53

There are two equations for the time, and the other is: $t = {v_0 cos(\alpha) \over g sin(\beta)}$

Here's why: We know that two lines are perpendicular if and only if product of their slope is minus one. The slope of the surface is $tan(\beta)$. Let the angle between velocity vector and the x axis be $\theta$, then $tan(\beta) tan(\theta) = -1$

$tan(\theta(t)) = {v_y(t) \over v_x(t)}$

$v_x(t) = v_x(0) = v_0 cos(\alpha + \beta)$

$v_y(t) = v_y(0) - gt = v_0 sin(\alpha + \beta) - gt$

$tan(\theta) = {-1 \over tan(\beta)} = {v_0 sin(\alpha + \beta) - gt \over v_0 cos(\alpha + \beta)}$

Solving fo $t$ yields: $t = {v_0 \over g}(sin(\alpha + \beta) + {cos(\alpha + \beta) \over tan(\beta)})$

Using trigonometric identities and after simplifying:

$t = {v_0 \over g}(sin(\beta)cos(\alpha) + {cos(\alpha)cos^2(\beta) \over sin(\beta)}) = {v_0 \over g}[{cos(\alpha) \over sin(\beta)} (sin^2(\beta) + cos^2(\beta))] = {v_0 cos(\alpha) \over g sin(\beta)}$

To verify the formula let $\beta$ goes to zero, then one normally expects that it would vertically only if either when it was launched vertically or when it's flight time approaches infinity. Which is consistent with this formula but not with yours.

To find the height of the strike point substitute the above values for $t$ and $v_y(0)$ into this equation: $h = v_y(0)t - {g t^2 \over 2}$

Edit: There are two independent equations for the time so there must be a constraint between $\alpha$ and $\beta$ which we obtain from equating these two : $tan(\alpha) = {cot(\beta) \over 2}$

The steps to derive the other equation for time is described in the above picture but I'm going to write it for completeness: First write velocities and accelerations in new coordinate frame along the surface: $v_{x'}(0) = v_0 cos(\alpha)$

$v_{y'}(0) = v_0 sin(\alpha)$

$a_{x'} = -g sin(\beta)$

$a_{y'} = -g cos(\beta)$

At the strike point $y'$ should be zero, so: $y'(t) = v_0 sin(\alpha) t - {1 \over 2}g cos(\beta)t^2 = 0$

$t = {2v_0 cos(\alpha) \over g sin(\beta)}$

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Okz. I'll undo it Just edit it a bit.So,both of the time period's are importantly equal for the particle to hit horizontally. –  ABC Apr 29 '13 at 17:16
Not exactly the range, it is height of that point from the ground, if I have interpreted correctly... –  Rohan R Apr 29 '13 at 17:19
@RohanR I checked your expression for the height and it's correct. Perhaps if you apply trigonometric identities and the constraint we have found $tan(\alpha) = {cot(\beta) \over 2}$ you'll arrive at your textbook formula. What your textbook got for the height? –  Azad Apr 29 '13 at 17:52
My textbook got the expression (2*u^2)/(g*(4+cot^2(beta))). Sorry about the weird formatting, idk how to use LaTeX. –  Rohan R Apr 30 '13 at 2:59