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3-2.Motion in Plane
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A ball is thrown from ground at an angle $\theta$ 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 $\theta$ but with a reduced speed of $u_0 / \alpha$. 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 $0.8 V _1$, the value of $\alpha$ is. . . . . .
A$2$
B$3$
C$4$
D$5$
(IIT-2019)
Solution
Average velocity $=\frac{\text { Total displacement }}{\text { Total time }}$
$\text { Total time taken }=t_1+t_2+t_3+\ldots . . . . . . . .$
$=t_1+\frac{t_1}{\alpha}+\frac{t_1}{\alpha^2}+\ldots \ldots . .$
$\text { Total time }=\frac{t_1}{1-\frac{1}{\alpha}}$
$\text { Total displacement }=v_1 t_1+v_2 t_2+\ldots \ldots . . .$
$=v_1 t_1+\frac{v_1}{\alpha} \cdot \frac{t_1}{\alpha}+\ldots \ldots . .$
$=\frac{v_1 t_1}{1-\frac{1}{\alpha^2}}$
On solving
$< v \rangle=\frac{ v _1 \alpha}{\alpha+1}=0.8 v _1$
$\alpha=4.00$
$\text { Total time taken }=t_1+t_2+t_3+\ldots . . . . . . . .$
$=t_1+\frac{t_1}{\alpha}+\frac{t_1}{\alpha^2}+\ldots \ldots . .$
$\text { Total time }=\frac{t_1}{1-\frac{1}{\alpha}}$
$\text { Total displacement }=v_1 t_1+v_2 t_2+\ldots \ldots . . .$
$=v_1 t_1+\frac{v_1}{\alpha} \cdot \frac{t_1}{\alpha}+\ldots \ldots . .$
$=\frac{v_1 t_1}{1-\frac{1}{\alpha^2}}$
On solving
$< v \rangle=\frac{ v _1 \alpha}{\alpha+1}=0.8 v _1$
$\alpha=4.00$
Standard 11
Physics
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