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JEE Advanced · Physics · 4. Motion in 2D

A ball is thrown from ground at an angle θ with horizontal and with an initial speed u0. 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. After hitting the ground, ball rebounds at the same angle θ but with a reduced speed of u0α. 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 V1, the value of α is______

  1. A 2
  2. B 4
  3. C 6
  4. D 8
Verified Solution

Answer & Solution

Correct Answer

(B) 4

Step-by-step Solution

Detailed explanation

For first projectile
Average velocity, \(\langle V\rangle=\frac{R}{T}=U_x=V_1\) \(\left[\begin{array}{c}(\text { where } R=\text { Range and } T=\text { Time of flight }) \\ R=\frac{2 U_x \cdot U_y}{g} \quad T=\frac{2 \cdot U_y}{g}\end{array}\right]\)
For journey
\(\langle V\rangle_{1 \rightarrow n}=\frac{R_1+R_2+\ldots+R_n}{T_1+T_2+\ldots+T_n}=\frac{\frac{2 u_{x_1} u_{y_1}}{g}+\frac{2 u_{x_2} u_{y_2}}{g}+\ldots+\frac{2 u_{x_n} u_{y_n}}{g}}{\frac{2 u_{y_1}}{g}+\frac{2 u_{y_2}}{g}+\ldots \frac{2 u_{y_n}}{g}}\)
\(\Rightarrow U_x\left[\frac{1+\frac{1}{\alpha^2}+\frac{1}{\alpha^4}+\ldots \frac{1}{\alpha^{2 n}}}{1+\frac{1}{\alpha}+\frac{1}{\alpha^2}+\ldots+\frac{1}{\alpha^n}}\right]=0.8~ v_1,\)\(\left[\begin{array}{c}1+\frac{1}{\alpha^2}+\frac{1}{\alpha^4}+\ldots \frac{1}{\alpha^{2 n}} \Rightarrow \text { Geometric progression } \\ \text { Sum of G.P. }=\frac{1}{1-\alpha^2}\end{array}\right]\)
\(\Rightarrow \frac{V_1\left[\frac{1}{1-\frac{1}{\alpha^2}}\right]}{\left[\frac{1}{1-\frac{1}{\alpha}}\right]}=0.8 v_1 \Rightarrow \frac{\alpha}{1+\alpha}=0.8 \Rightarrow \alpha=4\)
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