Jee Main 2024 29 01 2024 Shift 1 - Question12
Question 12
Consider a series of steps as shown. A ball is thrown from $O$. Find the minimum speed of directly jump to $5^{\text {th }}$ step.
(1) $5(\sqrt{2}+1) m / s$
(2) $5 \sqrt{2} m / s$
(3) $5 \sqrt{\sqrt{2}+1} m / s$
(4) $6 \sqrt{\sqrt{3}+1} m / s$
Show Answer
Answer: (3)
Solution:
$y=x \tan \theta-\frac{g x^{2}}{2 v^{2} \cos ^{2} \theta}$
$(2.5,2.5)$ must lie on this
$\Rightarrow 1=\tan \theta-\frac{g \times 2.5}{2 v^{2} \cos ^{2} \theta}$
$\Rightarrow \frac{25}{2 v^{2} \cos ^{2} \theta}=\tan \theta-1$
$\Rightarrow v^{2}=\frac{25}{2}{\frac{1+\tan ^{2} \theta}{\tan \theta-1} }$
$\Rightarrow v _{\text {min }}=5 \sqrt{\sqrt{2}+1}$
$[$ Happens when $\tan \theta=\sqrt{2}+1]$
BETA
