Centre of Mass 3 Question 19
22. A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of $120 m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is
(2018 Adv.)
Assertion and Reason
Mark your answer as
(a) If Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
(b) If Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I
(c) If Statement I is true; Statement II is false
(d) If Statement I is false; Statement II is true
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Answer:
Correct Answer: 22. (30)
Solution:
- $\because H=\frac{u^{2} \sin ^{2} 45^{\circ}}{2 g}=120 m \Rightarrow \frac{u^{2}}{4 g}=120 m$
If speed is $v$ after the first collision, then speed should remain $\frac{1}{\sqrt{2}}$ times, as kinetic energy has reduced to half.
$$ \begin{aligned} \Rightarrow \quad v & =\frac{u}{\sqrt{2}} \\ \therefore \quad h _{\max } & =\frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}=\frac{(u / \sqrt{2})^{2} \sin ^{2} 30^{\circ}}{2 g} \\ & =\frac{u^{2} / 4 g}{4}=\frac{120}{4}=30 \end{aligned} $$
