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Kinematics doubt...Plz help

Kinematics doubt...Plz help

by Sinjan Jana -
Number of replies: 2

Question:

A ball is thrown up with a certain velocity so that it reaches a height h. Find the ratio of times in which it is at h/3.


(A)\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)

(B)\( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)

(C)\( \frac{\sqrt{3}-1}{\sqrt{3}+1}\)

(D)\(\frac{1}{3}\)

I actually have done the problem like this..

There are two journeys.. one going up and going down

While going up

\(u = \sqrt{2gh}\)

So,

\(\frac{h}{3} ~=~\sqrt{2gh}~t_1~-~\frac{gt^2}{2}\)

Using,

\(\frac{-b~(+/-)~\sqrt{b^2-4ac}~}{2a}\)

I get

\( t_1=\frac{2\sqrt{2gh}~(+/-)\sqrt{8gh~-~\frac{8gh}{3}}}{2g}\)


Now, while coming down,

\(\frac{2h}{3}~=\frac{g(t_2)^2}{2}\)

Therefore,

\(t_2=(+/-)\sqrt{\frac{4h}{3g}}\)


After this I am totally stuck.. can you plz point out the mistakes I made in the above and tell me how to proceed with the calculation?

In reply to Sinjan Jana

Re: Kinematics doubt...Plz help

by Manjunath N -

dear sinjan,

initially

u=(2gh)^1/2

when the ball is at h/3

use v^2=u^2+2as

where

a= -g,s=h/3

we obtain v=(+/-)(4gh/3)^1/2

this implies ,

initially when the ball is thrown up

v= +(4gh/3)^1/2

on its downward journey

v= -(4gh/3)^1/2

now use v=u+at   or     t=(v-u)/a  when the ball is travelling both op & down

we obtain t1/t2 is the same as option 1 or option 2

(considering t1=time when it is at h/3 for the 1st time

& t2=time when it is at h/3 for the 2nd time)

In reply to Sinjan Jana

Re: Kinematics doubt...Please help

by Manish Verma -

Well, the equation \(y=u_yt+\frac{1}{2}a_yt^2\) can be used for both directions as y denotes the y coordinate of the position and the same y coordinate comes two times in the journey, once while going up and the other while coming down.

So, \(\frac{h}{3}=\sqrt{2gh}t-\frac{1}{2}gt^2\) itself can provide both t1 and t2 which are the roots of the quadratic.