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  • A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to of the height through which it falls. Find the average speed of the b

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to \frac{81}{100} of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms -2).
 
Option: 1 2.0\; ms^{-1}
Option: 2 3.50\; ms^{-1}
Option: 3 3.0\; ms^{-1}  
Option: 4 2.50\; ms^{-1}

Answers (1)

best_answer

If a ball is dropped from a height h on a horizontal floor, then it strikes with the floor with a speed.
v_{0}=\sqrt{2 g h_{0}} \quad\left[\text { From } v^{2}=u^{2}+2 g h\right]
\begin{aligned} &v_{1}=e v_{0}=e \sqrt{2 g h_{0}} \quad\left[\text { As } e=\frac{\text { velocity after collision }}{\text { velocity before collision }}\right]\\ &\text { (1) First height of rebound : } h_{1}=\frac{v_{1}^{2}}{2 g}=e^{2} h_{0}\\ &\therefore h_{1}=e^{2} h_{0} \end{aligned}

Height of the ball after nth rebound : Obviously, the velocity of ball after nth rebound will be
\begin{aligned} &v_{n}=e^{n} v_{0}\\ &\text { Therefore the height after nth rebound will be } \end{aligned}

\begin{array}{l} h_{n}=\frac{v_{n}^{2}}{2 g}=e^{2 n} h_{0} \\ \therefore h_{n}=e^{2 n} h_{0} \end{array}

Total distance travelled by the ball before it stops bouncing

\begin{array}{l} H=h_{0}+2 h_{1}+2 h_{2}+2 h_{3}+\ldots=h_{0}+2 e^{2} h_{0}+2 e^{4} h_{0}+2 e^{6} h_{0}+\ldots \\ H=h_{0}\left[1+2 e^{2}\left(1+e^{2}+e^{4}+e^{6} \ldots\right)\right] \\ =h_{0}\left[1+2 e^{2}\left(\frac{1}{1-e^{2}}\right)\right] \\ {\left[\text { As } 1+e^{2}+e^{4}+\ldots=\frac{1}{1-e^{2}}\right]} \\ \therefore H=h_{0}\left[\frac{1+e^{2}}{1-e^{2}}\right] \end{array}

Total time taken  by the ball to stop bouncing
\begin{array}{l} T=t_{0}+2 t_{1}+2 t_{2}+2 t_{3}+. .=\sqrt{\frac{2 h_{0}}{g}}+2 \sqrt{\frac{2 h_{1}}{g}}+2 \sqrt{\frac{2 h_{2}}{g}}+. \\ =\sqrt{\frac{2 h_{0}}{g}}\left[1+2 e+2 e^{2}+\ldots \ldots\right]\left[\text { As } h_{1}=e^{2} h_{0}: h_{2}=e^{4} h_{0}\right] \\ =\sqrt{\frac{2 h_{0}}{g}}\left[1+2 e\left(1+e+e^{2}+e^{3}+\ldots \ldots\right)\right] \\ =\sqrt{\frac{2 h_{0}}{g}}\left(\frac{1+e}{1-e}\right) \\ \therefore T=\left(\frac{1+e}{1-e}\right) \sqrt{\frac{2 h_{0}}{g}} \end{array}

SO by putting the value 'e', ho and g, we can get average velocity

So,
\mathrm{v}_{\mathrm{av}}=\frac{\mathrm{s}}{\mathrm{t}}=2.5 \mathrm{~m} / \mathrm{s}
 

Posted by

Deependra Verma

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