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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Advanced 2019 · Paper 2 · Physics · Motion in Two Dimensions

A ball is thrown from ground at an angle θ with horizontal and with...

A ball is thrown from ground at an angle θ\theta with horizontal and with an initial speed u0.{u}_{0} . For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is V1.{V}_{1} . After hitting the ground, ball rebounds at the same angle θ\theta but with a reduced speed of u0α.\dfrac{{u}_{0}}{\alpha} . Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is 0.8V1,0.8{V}_{1}, the value of α\alpha is______
A ball is thrown from ground at an angle θ with horizontal and with an initial speed u _ 0 . For the resulting

Correct answer

0

Step-by-step solution

For first projectile
Average velocity, \left(V\right) = \dfrac{R}{T} = {U}_{x} = {V}_{1}\left((whereR=RangeandT=Timeofflight)R=2Ux.UygT=2.Uyg\begin{array}{ll}\left(whereR = RangeandT = Timeofflight\right) \\ R = \dfrac{2{U}_{x} . {U}_{y}}{g} T = \dfrac{2 . {U}_{y}}{g}\end{array}\right)
For journey
(V)1→n=R1+R2+…+RnT1+T2+…+Tn=2ux1uy1g+2ux2uy2g+…+2uxnuyng2uy1g+2uy2g+…2uyng{\left(V\right)}_{1 \to n} = \dfrac{{R}_{1} + {R}_{2} + \ldots + {R}_{n}}{{T}_{1} + {T}_{2} + \ldots + {T}_{n}} = \dfrac{\dfrac{2{u}_{{x}_{1}}{u}_{{y}_{1}}}{g} + \dfrac{2{u}_{{x}_{2}}{u}_{{y}_{2}}}{g} + \ldots + \dfrac{2{u}_{{x}_{n}}{u}_{{y}_{n}}}{g}}{\dfrac{2{u}_{{y}_{1}}}{g} + \dfrac{2{u}_{{y}_{2}}}{g} + \ldots \dfrac{2{u}_{{y}_{n}}}{g}}
⇒ {U}_{x}\left(\dfrac{1 + \dfrac{1}{{\alpha}^{2}} + \dfrac{1}{{\alpha}^{4}} + \ldots \dfrac{1}{{\alpha}^{2n}}}{1 + \dfrac{1}{\alpha} + \dfrac{1}{{\alpha}^{2}} + \ldots + \dfrac{1}{{\alpha}^{n}}}\right) = 0.8{v}_{1}, \left(1+1α2+1α4+…1α2n⇒GeometricprogressionSumofG.P.=11−α2\begin{array}{ll}1 + \dfrac{1}{{\alpha}^{2}} + \dfrac{1}{{\alpha}^{4}} + \ldots \dfrac{1}{{\alpha}^{2n}} ⇒ Geometricprogression \\ SumofG . P . = \dfrac{1}{1 - {\alpha}^{2}}\end{array}\right)
⇒V1(11−1α2)(11−1α)=0.8v1⇒α1+α=0.8⇒α=4⇒ \dfrac{{V}_{1}\left(\dfrac{1}{1 - \dfrac{1}{{\alpha}^{2}}}\right)}{\left(\dfrac{1}{1 - \dfrac{1}{\alpha}}\right)} = 0.8{v}_{1} ⇒ \dfrac{\alpha}{1 + \alpha} = 0.8 ⇒ \alpha = 4
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