A particle is projected with velocity $v_0$ horizontally from a height $H$ over a hill, and strikes the hill perpendicularly.

Find $2gH/v_0 ^ 2$

Assume the hill rises linearly from the origin at an angle $\theta = \dfrac\pi6 = \si{30}^\circ$ .

The answer is 5.

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A unique and fabulous approach

Ujjwal Mani Tripathi
- 4 years, 11 months ago

This can be easily generalized. Doing that we get :-

$v_0 = \sqrt{\dfrac{2gh}{2 + \cot^2\theta} }$

Absolutely brilliant and unique approach!!

Pranav Saxena
- 4 years, 2 months ago

Same method...

Preetam Kandula
- 4 years, 11 months ago

That's Good! My main motive to post this question here was to know different methods to solve this problem, as the one I used is a bit time consuming. Sargam Yadav posted one good solution.

Vatsalya Tandon
- 4 years, 11 months ago

@Vatsalya Tandon try my problem That is close 2 , it also needs similar approach

Ujjwal Mani Tripathi
- 4 years, 11 months ago

Well this question is correct I solved it on my own... It has given all the required details...

Preetam Kandula
- 4 years, 11 months ago

how did you upload that picture ? because i used the same way of heights and equation of motion and i am getting the answer as 4 root 3 i would like you to check it.

A Former Brilliant Member
- 4 years, 11 months ago

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I use desktop. First open solutions. In that you have option to add picture. But if you are using an app then I don't know.

Preetam Kandula
- 4 years, 11 months ago

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i found a way to upload my solution you can go here : https://brilliant.org/problems/solve-it-17/ solve the problem and i have added my solution there . please check and tell me my mistake , i'll be glad

A Former Brilliant Member
- 4 years, 10 months ago

Well anyways just tell whats the difference between your solution and my solution. I mean time or v'.

Preetam Kandula
- 4 years, 11 months ago

Well everyone just try to solve this question https://brilliant.org/problems/confusing-sometimes/ posted by me

Preetam Kandula
- 4 years, 11 months ago

5 Helpful
0 Interesting
0 Brilliant
0 Confused

Did the same!!

Prakhar Bindal
- 4 years, 11 months ago

I also did the same.

Kushagra Sahni
- 4 years, 11 months ago

i did the same way too

Ujjwal Mani Tripathi
- 4 years, 11 months ago

Can anyone tell me how to attach a photo with a soln..

Sargam yadav
- 4 years, 11 months ago

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Hi Sargam. Click "Edit" on your solution. That will bring up some formatting options like this:

If you click the third button from the left, it will bring up a upload menu that lets you upload an image. It will then insert the link to your image directly into your solution which you can then move around as you wish.Log in to reply

But there are no formatting options coming...

Sargam yadav
- 4 years, 10 months ago

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@Sargam Yadav – Can you be more descriptive about what happens when you click edit?

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@Josh Silverman – When i click edit — (1)sargam yadav july 13,2016 (2)Relevent wiki (optional)[start typing to select a wiki] (3) [body of solution] (4) formatting guide (5) submit (6) cancel, preview..

Sargam yadav
- 4 years, 10 months ago

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@Sargam Yadav – Are you using a phone or a computer?

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@Josh Silverman – I am using a phone...

Sargam yadav
- 4 years, 10 months ago

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@Josh Silverman – no,there is. i read a same que on brilliant site which i asked. the ans was written that one have to first upload the photo to the apps such as flickr or some sites that they had mentioned(i do not remember) and then copy the image link to where you have to upload the photo. i tried using the app flickr ,then copy and paste but i failed to upload that photo at brilliant.......do not know what to do..... sorry for taking your so much precious time..... but its okay if there is not a way.... thank you so much......

Sargam yadav
- 4 years, 10 months ago

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But there are no formatting options coming..

Sargam yadav
- 4 years, 10 months ago

3 Helpful
0 Interesting
0 Brilliant
0 Confused

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Equation of trajectory is -

$\large y = x\tan\theta - \frac{gx^2}{2V_0 ^2cos^2 \theta}$

$\large y = - \frac{gx^2}{2V_0 ^2}$

$\large \frac{dy}{dx} = -\frac{gx}{V_0 ^2}$

This is trajectory is perpendicular to planck when it hits it.

$\therefore m_1m_2 = -1$

$\large x = \frac{\sqrt{3} V_0 ^2}{g}$

Also

$x = V_0 t$

$\large \therefore t = \sqrt{3} \frac{V_0 }{g}$

Now, $\large \tan(60) = \sqrt(3) = \frac{V_0 t}{H - 0.5gt^2}$

Equating, we get-

$\large \frac{\sqrt{3}gt^2}{2} + V_0 t - \sqrt{3}H = 0$

Plugging in the value of $t$ , we get-

$\large \frac{2gH}{V_0 ^2} = 5$