NSEP 2015 Problem 4 | Rotational Motion

A body released from a height H hits elastically an inclined plane at a point P. After the impact the body starts moving horizontally and hits the ground. The height at which point P should be situated so as to have the total time of travel maximum is

(a) H (b) 2H (c) $\frac{H}{4} (d) $\frac{H}{2}$

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(d) $\frac{H}{2}$

From the height H up to the point P, the ball will have a free fall. Hence, the time is fixed.

We just have to think how to get the path of the maximum time in the next part of the motion.

Hint: It can also be considered as free fall by only considering y direction.

The path is randomly drawn.

From P to ground, the height is h. Then, as the height is h, the time needed to fall is,

$$ t_{PA} = \sqrt{\frac{2x}{g}}$$

From B to P, the time needed is,

$$ t_{BP} = \sqrt{\frac{2(H-x)}{g}}$$

The total time is,

$$t = t_{BP}+t_{PA} = \sqrt{\frac{2(H-x)}{g}} +\sqrt{\frac{2x}{g}}$$

For maximum,

$$ \frac{dt}{dx}=0 $$

Calculating this,

$$ \frac{1}{2}\sqrt{\frac{2}{g}}\frac{-1}{\sqrt{H-x}} + \frac{1}{2}\sqrt{\frac{2}{g}}\frac{1}{\sqrt{x}} $$

solving this,

$$x=\frac{H}{2}$$

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