Question

A particle is projected upwards from level ground at an angle with horizontal such that its range is twice the maximum height reached by it.The angle made by its velocity vector with the horizontal at ahorizontal distance R/4 from the point of projection(R is horizontal range)

Answer 1

@yash2651995 plzz help!

Answer 2

@yash2651995 i posted this question already but did nt understand make u make me understnad

Answer 3

i'll do it in a systematic way.. BTW heena's way was right she found a connection.. and took out the answer..
if anyone else can also do it, it'd be appreciated.. as there are lots of ways to reach the answer specially in kinematics :)

Answer 4

OK Plz go on!

Answer 5

yea yash go on :)

Answer 6

I somehow got the first part of question's realtion. but final part not getting.

Answer 7

@heena nhi aya toh tum hi samjha diyo!!
after all you are an excellent teacher !! :P :D

Answer 8

\[V_r = (u\sin\theta)^2/2g\]
\[H_r = u t\cos \theta\]

Given that \[ 2V_r = H_r \]

Answer 9

ok

Answer 10

Sorry, \[ H_r = 2ut\cos \theta \]

Answer 11

Hr is Range??

Answer 12

v = u+at
\[ t = u\sin\theta/g\]

Answer 13

the qn is done or stilll need help @shakir

Answer 14

i need help

Answer 15

So we have
\[ (u \sin\theta)^2/2g = 2u\cos\theta u \sin\theta/g\]

\[ (u \sin\theta)^ = 4u^2\cos\theta \sin\theta\]
\[ \sin\theta = 4 \cos\theta \]

Answer 16

well @shakir @experimentX is helping u try to co-operate with him ok :) u ll undrstnd easily :)

Answer 17

ok

Answer 18

lol ... i am not sure ... check this
http://www.wolframalpha.com/input/?i=parametric+plot+x%3D2*30*cos%281.3258%29t+%2C+y+%3D+30*sin%281.3258%29t+-+1%2F2*9.8*t^2

Answer 19

I guess it worked
http://www.wolframalpha.com/input/?i=parametric+plot+x%3D2*25*cos%281.3258%29t+%2C+y+%3D+25*sin%281.3258%29t+-+1%2F2*9.8*t^2

Answer 20

so basically you have \( \theta = 75.96 \degree \) which is the initial angle of projection

Answer 21

The answer should be degree

Answer 22

45 degree

Answer 23

Wat's the final answer?

Answer 24

i guess this is already solved @shakir
http://openstudy.com/study#/updates/4fc4c3d8e4b0964abc8718ab

Answer 25

neh neh, that was some other question's answer.. my workbook is a mess..

Answer 26

now you have relation ...
\[ y(x) = u t \sin(75.96) - \frac 12 gt^2 , \; \; x(t) = u\cos(75.96) t\]
The horizontal range is given by relation
\[
H_r = 2u^2\cos(75.96)\sin(75.96)/g\]

Find, dy/dx = dy/dt * 1/(dx/dt)| H_r/4

tan inverse of this value should be your answer ...

Answer 27

\[ t = \frac x{u \cos(75.96)}\]
So we have
\[ y = x \tan(75.96) - \frac {g x^2}{2 u^2 \cos^2(75.96) } \]
\[ \frac{dy}{dx} = \tan(75.96) - \frac {g x}{u^2 \cos^2(75.96) }, x = Hr/4=2u^2\cos(75.96)\sin(75.96)/4g = \\
u^2\cos(75.96)\sin(75.96)/2g\]

\[ =\tan(75.96) - \frac{gu^2\cos(75.96)\sin(75.96)/2g}{u^2 \cos^2(75.96)} \]

\[ =\tan(75.96) - \frac{\sin(75.96)/2}{\cos(75.96)} \]

Looks like i didn't get 45

Answer 28

i'll check and reply back

Answer 29

http://www.wolframalpha.com/input/?i=d%2Fdx+%2825*sin%281.3258%29%28x%2F%282*25*cos%281.3258%29%29%29+-+1%2F2*9.8*%28x%2F%282*25*cos%281.3258%29%29%29^2%29+%2C+x%3D60%2F4
Ah it's around 45 degrees

Answer 30

that too lengthy method @experimentX we can also use it like this way
R=2hmax
if R get 1/4 means 2Hmax will also get 1/4
R/4=Hmax/2
we know
Hmax=u^2/2g
R/4=u^2/4g [R=u^2sin2theta/g and Hmax=u^2/2g] u^2sin2theta/4g=u^2/4g sin2theta=1
2theta=90
theta=90/2
theta=45

Answer 31

or there is one more way

R=2hmax if R get 1/4 means 2Hmax will also get 1/4
R/4=Hmax/2
we know
Hmax=u^2/2g R/4=u^2/4g
here U=initial velcoity R=u^2/2g
this is the formula of max range and it comes in 45degree

Answer 32

srry for wrong statement its
here U=initial velcoity R=u^2/g

Answer 33

All messed up in equation ... :(
best way is to model a parabola
y = ax^2 + bx + c
put some points (-2,0), (0,2), (2,0)
solve for a,b,c
http://www.wolframalpha.com/input/?i=solve+0%3D4a-2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc
find the slope at R/4
http://www.wolframalpha.com/input/?i=solve+0%3D4a-2b%2Bc%2C+0%3D4a%2B2b%2Bc%2C+2%3Dc
take tan inverse which is pi/4

Answer 34

mathematics seems a lot simpler than physics!!

Answer 35

it might be but i m not a studnt of maths so dunno much abu dat :)