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# NSEP 2015 Problem 4 | Rotational Motion Try out this problem on the one dimensional motion from National Standard Examination in Physics 2015-2016.

## NSEP 2015-16 ~ Problem 4

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

### Key Concepts

Freely falling body

Time needed for a free body

## Suggested Book | Source | Answer

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) $\frac{H}{2}$

## Try with Hints

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. B is the point at height H. From P the ball goes horizontally. Then due to gravity it has some projectile motion.

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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