Question 160: Work, Energy and Power & Circular Motion

Question

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 $ 	o $ $\infty $ is :

Accepted Answer

Let,
Kinetic energy (k) = ${1 \over 2}$ m $\upsilon $
2
before it hit the ground.
After hitting the ground kinetic energy
(k') = ${1 \over 2}$ m $\upsilon $$_1^2$
$	herefore\,\,\,$According to the question,
${1 \over 2}$ m$\upsilon $$_1^2$ = ${1 \over 2}$ $ 	imes $ ${1 \over 2}$ m$\upsilon $
2
$ \Rightarrow $$\,\,\,$ $\upsilon $
1
= ${v \over {\sqrt 2 }}$
After hitting the ground the object will bounce
h' =  ${{v_1^2} \over {2g}}$ = ${{{v^2}} \over {4g}}$ = ${h \over 2}$ [ as    h = ${{{v^2}} \over {2g}}$ ]
Total distance travelled from the time it first hits the ground to the next time it hits the ground is = ${h \over 2}$ + ${h \over 2}$ = h
So, this will create a infinite geometric progression with the common ration ${1 \over 2}$.
$	herefore\,\,\,$  Total distance covered
= h (distance travelled by the obhect when first dropped, before it hits the ground)
+ (h + ${h \over 2}$ + ${h \over 4}$ + . . . . . . . .$ \propto $)
= h + ${h \over {1 - {1 \over 2}}}$
= h + 2h
= 3h

Answer

3h

Answer

$\infty $

Answer

${5 \over 3}$h

Answer

${8 \over 3}$h

Work, Energy and Power & Circular Motion

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 \to \infty is :

Choose the correct answer: