JEE Main 2019 · Physics · Kinematics and Laws of Motion · hard

A ball is dropped from a height of 20 m onto a hard horizontal floor. At each impact the speed just after the bounce is half the speed just before it…

Single correct option

A ball is dropped from a height of

20\ \mathrm{m}

onto a hard horizontal floor. At each impact the speed just after the bounce is half the speed just before it (coefficient of restitution

0.5

). Neglecting air resistance, the total distance travelled by the ball before it finally comes to rest is

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Answer: (D) $\dfrac{100}{3}\ \mathrm{m}$

The rebound speed is

e

times the impact speed, and since height varies as speed squared, each rebound height is

e^2=\tfrac14

of the previous drop height. The first fall is

20\ \mathrm{m}

. Every later bounce contributes an upward and an equal downward path: heights

20e^2,\ 20e^4,\dots

, each counted twice. Total distance

=20+2\cdot20\,\dfrac{e^2}{1-e^2}=20+40\cdot\dfrac{1/4}{3/4}=20+\dfrac{40}{3}=\dfrac{100}{3}\ \mathrm{m}\approx33.3\ \mathrm{m}

. Using

e

instead of

e^2

in the series gives the incorrect

60\ \mathrm{m}

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake