The speed of a projectile at its maximum heig | vedclass

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

Question

(C) Let the initial speed be $u$ and the angle of projection be $	heta$.The speed at maximum height is $u \cos 	heta$.Given that $u \cos 	heta = \f

Vedclass · Accepted Answer

$4\sqrt{3}$

Vedclass · Answer

(C) Let the initial speed be $u$ and the angle of projection be $	heta$.The speed at maximum height is $u \cos 	heta$.Given that $u \cos 	heta = \frac{\sqrt{3}}{2} u$,so $\cos 	heta = \frac{\sqrt{3}}{2}$,which implies $	heta = 30^{\circ}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$ and the range $R$ is given by $R = \frac{u^2 \sin 2	heta}{g}$.We are given $R = P 	imes H$,so $P = \frac{R}{H}$.Substituting the formulas: $P = \frac{u^2 (2 \sin 	heta \cos 	heta) / g}{u^2 \sin^2 	heta / (2g)} = \frac{4 \cos 	heta}{\sin 	heta} = 4 \cot 	heta$.For $	heta = 30^{\circ}$,$P = 4 \cot 30^{\circ} = 4 	imes \sqrt{3} = 4\sqrt{3}$.

The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$ times its initial speed. If the range of the projectile is $P$ times the maximum height attained by it,$P$ is equal to

Similar Questions

If the potential energy of a projectile at its maximum height is $3/4$ of its initial kinetic energy,then the angle of projection is ...... $^o$.

$A$ particle of mass $m$ is projected with velocity $v$ making an angle of $45^{\circ}$ with the horizontal. When the particle lands on the level ground,the magnitude of the change in its momentum will be

$A$ projectile with speed $50 \ m \ s^{-1}$ is thrown at an angle of $60^{\circ}$ with the horizontal. The maximum height that can be reached is,(acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $m$)

The maximum height attained by a projectile is increased by $10 \%$ by keeping the angle of projection constant. What is the percentage increase in the time of flight (in $\%$)?

$A$ projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40\,m/s$. The velocity of the projectile at $t = 2\,s$ from the start will be $........$ (Given $g = 10\,m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module