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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 distance covered as $t\rightarrow \infty$ is :

Option 1)
3h

Option 2)
$\infty$

Option 3)
$\frac{5}{3}h$

Option 4)
$\frac{8}{3}h$

Answers (3)

best_answer

As we have learned

3rd equation or Velocity –displacement equation -

$V^{2}-u^{2}=2as$

$V\rightarrow$ Final Velocity

$s\rightarrow$ Displacement

$u\rightarrow$ Initial velocity

$a\rightarrow$ acceleration

-

Initial height covered $h_0 = h$

At the point of impact , velocity

$v_0 = \sqrt{2gh}$

After losing its kinetic energy

by 50 % velocity $v_1 = 1/2 v_0 = \sqrt {gh }$

$\Rightarrow h_1 = \frac{v_{1}^{2}}{2g } = h /2 \Rightarrow$ Tota l distance after 1st collision = $2 h_1 = h$

Similarly $h_2 = h /4 , h_3 = h/8$

Total distance covered

$= h + 2h_1 + 2 h_2 + .... \infty \\ = h + h + h/2 ...\infty \\ = 2h + h/2 (1+ 1/2 ...\infty )$

$= 2 h + h/2 (1-1/2 ) = 3 h$

Option 1)

3h

Option 2)

$\infty$

Option 3)

$\frac{5}{3}h$

Option 4)

$\frac{8}{3}h$

Posted by

SudhirSol

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2

Posted by

vishal dange

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First is right answer, because every time it hit ground it cannot much distance

Posted by

Omkar Kadam

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