Question

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

Answer 1

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.

Answer 2

Answer:

0.5seconds

height=h

(H-h)=1/2 gt2

substitute 2 with H-h

H-h/2+√2g/h

H/h=1/2divide you will get 0.5 as answer please make me brainlist