Try out this problem on the one dimensional motion from National Standard Examination in Physics 2015-2016.
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) $\frac{H}{2}$
Freely falling body
Time needed for a free body
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,
$$x=\frac{H}{2}$$
Physics Olympiad Program at Cheenta
Try out this problem on the one dimensional motion from National Standard Examination in Physics 2015-2016.
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) $\frac{H}{2}$
Freely falling body
Time needed for a free body
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,
$$x=\frac{H}{2}$$
Physics Olympiad Program at Cheenta
