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# An object is dropped from a height h from the ground.  Every time it hits the ground it looses 50% of its kinetic energy.  The total distanc

• Option 1)

3h

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• Option 2)

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• Option 3)

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• Option 4)

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Option:(1) is correct.
When particle starts from 'h' height and hits ground it travels 'h' distance.
Now its kinetic energy becomes half:
kinectic energy just before hitting the ground:
$2gh=1/2mv^2$
after hitting it becomes half:
$1/2(1/2mv^2)$
after hitting;
$2gx=1/2(1/2mv^2)$
$2gx=1/2(2gh)$              $\left [ 2gh=1/2mv^2 \right ]$
$x=h/2$
after hitting it travels $h/2$ distance upward and $h/2$ distance downward.
after next hitting it travels $h/4$ distance upward and $h/4$ disatnce downward
& this repeats continuously.
So, total distance travelled by particle is:
$h+h/2+h/2+h/4+h/4+h/8+h/8\cdots \infty$
$h+h+h/2+h/4+\cdots \infty$
Sum of GP=
$a/(1-r)$                 $a=h,r=1/2$
$h+ \left \{ h/\left ( 1-1/2 \right )\right \}$
$h+2h=3h$

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