Question 150: 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: