A particle is projected from the ground with | vedclass

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

Question

(C) Average velocity is defined as the ratio of displacement to time.Let the point of projection be $(0, 0)$. The highest point of the trajectory is a

Vedclass · Accepted Answer

$\frac{v}{2}\sqrt{1 + 3\cos^{2}	heta}$

Vedclass · Answer

(C) Average velocity is defined as the ratio of displacement to time.Let the point of projection be $(0, 0)$. The highest point of the trajectory is at coordinates $(\frac{R}{2}, H)$,where $R$ is the horizontal range and $H$ is the maximum height.The displacement vector $\vec{s}$ from the origin to the highest point is $\vec{s} = \frac{R}{2}\hat{i} + H\hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{(\frac{R}{2})^2 + H^2}$.The time taken to reach the highest point is $t = \frac{T}{2} = \frac{v\sin	heta}{g}$.We know $H = \frac{v^2\sin^2	heta}{2g}$ and $R = \frac{v^2\sin2	heta}{g} = \frac{2v^2\sin	heta\cos	heta}{g}$.Thus,$\frac{R}{2} = \frac{v^2\sin	heta\cos	heta}{g}$.$|\vec{s}| = \sqrt{(\frac{v^2\sin	heta\cos	heta}{g})^2 + (\frac{v^2\sin^2	heta}{2g})^2} = \frac{v^2\sin	heta}{g} \sqrt{\cos^2	heta + \frac{\sin^2	heta}{4}} = \frac{v^2\sin	heta}{2g} \sqrt{4\cos^2	heta + \sin^2	heta}$.Using $\sin^2	heta = 1 - \cos^2	heta$,we get $|\vec{s}| = \frac{v^2\sin	heta}{2g} \sqrt{3\cos^2	heta + 1}$.Average velocity $v_{av} = \frac{|\vec{s}|}{t} = \frac{\frac{v^2\sin	heta}{2g} \sqrt{3\cos^2	heta + 1}}{\frac{v\sin	heta}{g}} = \frac{v}{2} \sqrt{1 + 3\cos^2	heta}$.

$A$ particle is projected from the ground with an initial speed $v$ at an angle $\theta$ with the horizontal. The average velocity of the particle between its point of projection and the highest point of its trajectory is

Similar Questions

$A$ particle has an initial velocity of $(3\hat i + 4\hat j) \; ms^{-1}$ and an acceleration of $(0.4\hat i + 0.3\hat j) \; ms^{-2}$. Its speed after $10 \; s$ is:

$A$ projectile of mass $1 \, kg$ is projected with a velocity of $\sqrt{20} \, m/s$ such that it strikes on the same level as the point of projection at a distance of $\sqrt{3} \, m$. Which of the following options are incorrect:

An object of mass $m$ is projected with a momentum $p$ at such an angle that its maximum height is $\frac{1}{4}$th of its horizontal range. Its minimum kinetic energy in its path will be

$A$ particle has an initial velocity of $(\hat{i} + \hat{j}) \, m/s$ and an acceleration of $(\hat{i} + \hat{j}) \, m/s^2$. Its speed after $10 \, s$ will be:

Two balls are thrown simultaneously from the ground with the same velocity of $10\,m/s$ but at different angles of projection with the horizontal. Both balls fall at the same distance $5\sqrt{3}\,m$ from the point of projection. What is the time interval between the balls striking the ground?

Vedclass Test Series

Exam Paper Generator

Online Exam Module