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(D) Let $v$ be the velocity of the particle at the top of the incline. The particle performs projectile motion to cross the $40 \, m$ wide well at an

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$20\sqrt{2} \, m/s$

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(D) Let $v$ be the velocity of the particle at the top of the incline. The particle performs projectile motion to cross the $40 \, m$ wide well at an angle of $45^o$. The range $R$ of the projectile is given by $R = \frac{v^2 \sin(2	heta)}{g}$.Given $R = 40 \, m$ and $	heta = 45^o$,we have $40 = \frac{v^2 \sin(90^o)}{g} = \frac{v^2}{g}$.Assuming $g = 10 \, m/s^2$,we get $v^2 = 400$,so $v = 20 \, m/s$.Now,consider the motion on the frictionless incline of length $s = 20\sqrt{2} \, m$. The acceleration along the incline is $a = -g \sin(45^o) = -\frac{g}{\sqrt{2}}$.Using the equation of motion $v^2 = u^2 + 2as$,where $u$ is the initial velocity at point $M$:$(20)^2 = u^2 + 2(-\frac{g}{\sqrt{2}})(20\sqrt{2})$.$400 = u^2 - 2g(20) = u^2 - 40(10) = u^2 - 400$.$u^2 = 800$,which gives $u = \sqrt{800} = 20\sqrt{2} \, m/s$.

$A$ particle is projected with velocity $u$ from point $M$ on a frictionless incline of length $20\sqrt{2} \, m$. If it crosses a $40 \, m$ wide well at an angle of $45^o$,what should be its velocity at point $M$?

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