QJEEM17Phe622: JEE Main 2017 Physics - Class 11 JEEM17Phe622 Solution: An object is dropped from a height h from the ground. Every…

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 $\upsilon$ $_1^2$

$\therefore\,\,\,$ According to the question,

${1 \over 2}$ m $\upsilon$ $_1^2$ = ${1 \over 2}$ $\times$ ${1 \over 2}$ 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 ${1 \over 2}$ .

$\therefore\,\,\,$ 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