Tuesday, September 16, 2008

Q) A body falling from a given height H hits an inclined plane in its path at a height h . As a result of this impact, the direction of the velocity of the body becomes horizontal. Find the value of h/H for which the body will take the maximum time to reach the ground.

When it collides with the inclined plane at height h, the body's velocity is sqrt(2g(H-h) ).
And time taken to cover this distance is sqrt(2(H-h) / g).
(By using ---> s = (1/2) g.t^2 )

Now this becomes the horizontal velocity. So now time taken to reach the ground will be :
t = sqrt (2h/g) since downward velocity initially now is 0.

So total time taken = sqrt(2(H-h) / g) + sqrt (2h/g)
Time taken will be max when the two terms are equal (symmetry), so 2(H-h) = 2h
2H = 4h
h/H = 1/2