A ball is thrown from ground at angle with horizontal and with an initial speed . 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 . [2019]

Question

A ball is thrown from ground at 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, the ball rebounds at the same angle $θ$ but with a reduced speed of $\frac{u_{0}}{a}$ . 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 _____ [2019]

Smart Achievers · Accepted Answer

(4)

Let $u_{0}$ be the horizontal velocity of the projectiles and $t_{1}, t_{2}, t_{3}, \dots$ be the time taken as shown in figure.

$Average velocity = \frac{total displacement}{total time taken}$

$Time of flight T = \frac{2 u \sin θ}{g}$

$For given value of θ, value of T will change with the value of u .$

$∴$ $Total time taken = t_{1} + t_{2} + t_{3} + \dots$

$= t_{1} + \frac{t_{1}}{α} + \frac{t_{1}}{α^{2}} + \dots = \frac{t_{1}}{1 - \frac{1}{α}} = \frac{t_{1} α}{α - 1}$

$Total displacement = u_{1} t_{1} + u_{2} t_{2} + u_{3} t_{3} + \dots$

$= u_{1} t_{1} + \frac{u_{1}}{α} \cdot \frac{t_{1}}{α} + \frac{u_{1}}{α^{2}} \cdot \frac{t_{1}}{α^{2}} + \dots$ $= \frac{u_{1} t_{1}}{1 - \frac{1}{α^{2}}} = \frac{u_{1} t_{1} α^{2}}{α^{2} - 1}$

$∴$ $Average velocity = \frac{total displacement}{total time}$

$= \frac{u_{1} t_{1} α^{2}}{α^{2} - 1} \times \frac{α - 1}{t_{1} α} = \frac{u_{1} α}{α + 1}$

$According to question, average velocity = 0.8 V_{1}$

$\Rightarrow 0.8 V_{1} = \frac{u_{1} α}{α + 1} . . . (i)$

$V_{1} = \frac{u_{1} t_{1}}{t_{1}} = u_{1} . . . (i i)$

$From (i) and (ii):$

$α = 0.8 α + 0.8 or α - 0.8 α = 0.8 or 0.2 α = 0.8$

$or α = \frac{0.8}{0.2} = 4$

Solution

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