A particle is projected with velocity 2 sqrt{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the velocity of projection be $v = 2\sqrt{gh}$ at an angle $	heta$ with the horizontal.Horizontal component of velocity: $v_x = v \cos 	heta

Vedclass · Accepted Answer

$\sqrt{\frac{4h}{g}}$

Vedclass · Answer

(A) Let the velocity of projection be $v = 2\sqrt{gh}$ at an angle $	heta$ with the horizontal.Horizontal component of velocity: $v_x = v \cos 	heta = 2\sqrt{gh} \cos 	heta$.The time taken to cover the horizontal distance $d = 2h$ between the walls is $t = \frac{d}{v_x} = \frac{2h}{2\sqrt{gh} \cos 	heta} = \sqrt{\frac{h}{g}} \sec 	heta$ ... $(i)$Using the equation of trajectory $y = x 	an 	heta - \frac{gx^2}{2v^2 \cos^2 	heta}$,for the two walls at $x_1$ and $x_2$ where $y = h$:$h = x 	an 	heta - \frac{gx^2}{2(4gh) \cos^2 	heta} \Rightarrow h = x 	an 	heta - \frac{x^2}{8h \cos^2 	heta}$.Rearranging gives $x^2 - (8h 	an 	heta \cos^2 	heta) x + 8h^2 = 0$.The distance between the walls is $x_2 - x_1 = 2h$. For a quadratic $ax^2 + bx + c = 0$,$|x_2 - x_1| = \frac{\sqrt{D}}{|a|}$.$|x_2 - x_1| = \frac{\sqrt{(8h 	an 	heta \cos^2 	heta)^2 - 4(8h^2)}}{1} = 2h$.$64h^2 	an^2 	heta \cos^4 	heta - 32h^2 = 4h^2 \Rightarrow 16 \sin^2 	heta \cos^2 	heta - 8 = 1 \Rightarrow 4 \sin^2(2	heta) = 9$ (This implies a specific geometry).Alternatively,using the time of flight between walls $t = \frac{2v_y}{g}$,where $v_y$ is the vertical velocity at height $h$: $v_y^2 = (v \sin 	heta)^2 - 2gh = 4gh \sin^2 	heta - 2gh = 2gh(2\sin^2 	heta - 1)$.$t = \frac{2\sqrt{2gh(2\sin^2 	heta - 1)}}{g} = 2\sqrt{\frac{2h}{g}(2\sin^2 	heta - 1)}$.Equating the two expressions for $t$: $\sqrt{\frac{h}{g}} \sec 	heta = 2\sqrt{\frac{2h}{g}(2\sin^2 	heta - 1)}$.$\sec^2 	heta = 8(2\sin^2 	heta - 1) = 16\sin^2 	heta - 8$.$1 + 	an^2 	heta = 16\sin^2 	heta - 8 \Rightarrow 	an^2 	heta = 3 \Rightarrow 	heta = 60^\circ$.Substituting $	heta = 60^\circ$ into $(i)$: $t = \sqrt{\frac{h}{g}} \sec 60^\circ = 2\sqrt{\frac{h}{g}} = \sqrt{\frac{4h}{g}}$.

$A$ particle is projected with velocity $2 \sqrt{gh}$,so that it just flies over two walls of equal height $h$ and $2h$ distance apart from each other. Find the time for which the particle flies between the walls.

Similar Questions

The velocity at the maximum height of a projectile is $\frac{\sqrt{3}}{2}$ times its initial velocity of projection $(u)$. Its range on the horizontal plane is .............

$A$ particle is projected from the ground with a kinetic energy $E$ at an angle of $60^{\circ}$ with the horizontal. Its kinetic energy at the highest point of its motion will be

$A$ gun is aimed at a target in the line of its barrel. The target is released and allowed to fall under gravity at the same instant the gun is fired. The bullet will

Vedclass Test Series

Exam Paper Generator

Online Exam Module