A particle is projected from level ground. It | vedclass

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

Question

(C) The kinetic energy of a projectile is maximum at the point of projection and at the ground level (where speed is $v$),and minimum at the highest p

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) The kinetic energy of a projectile is maximum at the point of projection and at the ground level (where speed is $v$),and minimum at the highest point (where speed is $v \cos 	heta$).Given $\frac{K_{\max}}{K_{\min}} = 9$,we have $\frac{\frac{1}{2}mv^2}{\frac{1}{2}mv^2 \cos^2 	heta} = 9$.This simplifies to $\frac{1}{\cos^2 	heta} = 9$,which means $\cos^2 	heta = \frac{1}{9}$,so $\cos 	heta = \frac{1}{3}$.Since $\cos 	heta = \frac{1}{3}$,we find $\sin 	heta = \sqrt{1 - (\frac{1}{3})^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{2\sqrt{2}/3}{1/3} = 2\sqrt{2}$.The range $R$ is given by $R = \frac{v^2 \sin 2	heta}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$ and the maximum height $H$ is $H = \frac{v^2 \sin^2 	heta}{2g}$.The ratio $\frac{R}{H} = \frac{2v^2 \sin 	heta \cos 	heta / g}{v^2 \sin^2 	heta / 2g} = 4 \cot 	heta$.Substituting $\cot 	heta = \frac{1}{	an 	heta} = \frac{1}{2\sqrt{2}}$,we get $\frac{R}{H} = 4 	imes \frac{1}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

$A$ particle is projected from level ground. Its kinetic energy $K$ changes due to gravity such that $\frac{K_{\max}}{K_{\min}} = 9$. The ratio of the range $R$ to the maximum height $H$ attained during its flight is:

Similar Questions

The maximum horizontal range of a projectile is $160 \ m$. When the projectile is thrown with the same speed at an elevation of $30^{\circ}$ from the horizontal,it will reach a maximum height of ......... $m$.

$A$ stone is just released from the window of a train moving along a horizontal straight track. The stone will hit the ground following:

$A$ swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the edge at the bottom,given that the ledge at the bottom is $2 \ m$ wide and $10 \ m$ below the top of the cliff?

If the instantaneous velocity of a particle projected as shown in the figure is given by $v = a \hat{i} + (b - ct) \hat{j}$,where $a, b$,and $c$ are positive constants,the range on the horizontal plane will be

The equation for the trajectory of a projectile is $y = \left( \frac{x}{\sqrt{3}} - \frac{x^2}{60} \right) \text{ m}$. The velocity of projection of the projectile is (Acceleration due to gravity $g = 10 \text{ m s}^{-2}$) (in $\text{ m s}^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module