A) \[\frac{h}{2}\frac{\left[ 1-{{e}^{2}} \right]}{\left[ 1+{{e}^{2}} \right]}\]
B) \[\frac{h\left[ 1-{{e}^{2}} \right]}{\left[ 1+{{e}^{2}} \right]}\]
C) \[\frac{h}{2}\frac{\left[ 1+{{e}^{2}} \right]}{\left[ 1-{{e}^{2}} \right]}\]
D) \[\frac{h\left[ 1+{{e}^{2}} \right]}{\left[ 1-{{e}^{2}} \right]}\]
Correct Answer: D
Solution :
[d] The velocity of particle after falling through height h \[u=\sqrt{2gh}\] ...(i) After first rebounding, the velocity of ball is eu and after attaining maximum height it will come to the ground with same velocity eu. So, after second rebounding its velocity will be\[{{e}^{2}}u\]. Similarly, after third fourth...etc reboundings its velocities will be\[{{e}^{2}}u,\text{ }{{e}^{4}}u\],... etc. Since, it first rebounds with velocity eu so if it attains height h then from \[{{v}^{2}}={{u}^{2}}-2gh\] \[\therefore 0={{e}^{2}}{{u}^{2}}-2gh\] Or \[{{h}_{1}}=\frac{{{e}^{2}}{{u}^{2}}}{2g}=\frac{{{e}^{2}}2gh}{2g}={{e}^{2}}h\][from Eq. (i)] The same height the ball travels while approaching ground. Now, it rebounds with velocity \[{{e}^{2}}u\]so if it attains a height \[{{h}_{2}}\]then \[0 = {{e}^{4}}{{u}^{2}}- 2g{{h}_{2}}\] or \[{{h}_{2}}={{e}^{4}}h\] The similar process will follow for further reboundings Hence, the total distance travelled by the practice before it stops rebounding. \[=h+2{{h}_{i}}+2{{h}_{2}}+...\infty =h+2{{e}^{2}}h+2{{e}^{4}}h+...\infty \] \[=h+2{{e}^{2}}h\left( 1+{{e}^{2}}+{{e}^{4}}+...\infty \right)=h+2{{e}^{2}}h\left( \frac{1}{1-{{e}^{2}}} \right)\]\[=h\left( 1+\frac{2{{e}^{2}}}{1-{{e}^{2}}} \right)=\left( \frac{1+{{e}^{2}}}{1-{{e}^{2}}} \right)h\]
You need to login to perform this action.
You will be redirected in
3 sec
