A ball is thrown from ground at an angle \(\theta\) with horizontal and with an initial speed \(\mathrm{u}_{0}\). 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 \(\mathrm{V}_{1}\). After hitting the ground, ball rebounds at the same angle \(\theta\) but with a reduced speed of \(\mathrm{u}_{0} / \alpha\). 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 \(0.8 \mathrm{~V}_{1}\), the value of \(\alpha\) is ____

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Kinematics

A ball is thrown from ground at an angle

θ

with horizontal and with an initial speed

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

V_{1}

. After hitting the ground, ball rebounds at the same angle

θ

but with a reduced speed of

u_{0} / α

. 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

0.8 V_{1}

, the value of

α

is ____

Answer: 4

Total time taken:

T = t_{1} + \frac{t _{1}}{α} + \frac{t _{1}}{α ^{2}} + \dots = \frac{t _{1}}{1 - 1/ α}

Total displacement:

S = v_{1} t_{1} + \frac{v _{1}}{α} \cdot \frac{t _{1}}{α} + \dots = \frac{v _{1} t _{1}}{1 - 1/ α ^{2}}

Average velocity:

⟨ v ⟩ = \frac{S}{T} = \frac{v _{1} t _{1} / ( 1 - 1/ α ^{2} )}{t _{1} / ( 1 - 1/ α )} = \frac{v _{1} α}{α + 1}

Given

⟨ v ⟩ = 0.8 v_{1}

\frac{v _{1} α}{α + 1} = 0.8 v_{1}

α = 0.8 (α + 1)

0.2 α = 0.8

α = 4.00

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Numerical answer