A body falling freely from a given height h

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A body falling freely from a given height H hits an inclined plane on its path at a height h.As a result of this impact, the direction of the velocity of the body becomes horizontal. The body will take maximum time to reach the ground if

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Answer:

The body will take maximum time to reach the ground id h : H = 1 : 2

Explanation:

The body will travel a total height of H-h before striking the horizontal plane

Using the second equation of motion

H-h=0\times t+\frac{1}{2}gt^2

\implies H-h=\frac{1}{2}gt^2

\implies t=\sqrt{\frac{2(H-h)}{g}}

After striking the plane, the vertical componenet of the velocity will be zero and it will have only the horizontal component. The body will behave like a projectile, thrown horizontally

Time taken to strike the ground is t' then

h=0\times t'+\frac{1}{2}gt'^2

or, t'=\sqrt{\frac{2h}{g}}

Total time taken by the body

T = t + t'

T=\sqrt{\frac{2(H-h)}{g}}+\sqrt{\frac{2h}{g}}

For maxima

dT/dh = 0

or, \frac{1}{2} \Bigr[{\frac{2(H-h)}{g}}\Bigr]^{-1/2}\times(\frac{-2}{g})+\frac{1}{2}\Bigr[{\frac{2h}{g}}\Bigr]^{-1/2}\times(2/g)=0

\implies \Bigr[{\frac{2(H-h)}{g}}\Bigr]^{-1/2}=\Bigr[{\frac{2h}{g}}\Bigr]^{-1/2}

\implies {\frac{2(H-h)}{g}}={\frac{2h}{g}}

\implies 2(H-h)=2h

\implies H=2h

\implies H/h=2/1

Therefore the body will take maximum time to reach the ground id h : H = 1 : 2

Hope this answer is helpful.


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