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) The average velocity is defined as the total displacement divided by the total time taken.Let the particle be projected from the origin $(0,0)$. A

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The average velocity is defined as the total displacement divided by the total time taken.Let the particle be projected from the origin $(0,0)$. At the highest point, the coordinates are $(R/2, H)$, where $R$ is the horizontal range and $H$ is the maximum height.Displacement vector $\vec{s} = \frac{R}{2} \hat{i} + H \hat{j}$.The magnitude of displacement is $|\vec{s}| = \sqrt{(R/2)^2 + H^2}$.The time taken to reach the highest point is $t = \frac{T}{2} = \frac{v \sin 	heta}{g}$.We know $R = \frac{v^2 \sin 2	heta}{g} = \frac{2v^2 \sin 	heta \cos 	heta}{g}$ and $H = \frac{v^2 \sin^2 	heta}{2g}$.Thus, $R/2 = \frac{v^2 \sin 	heta \cos 	heta}{g}$.$|\vec{s}| = \sqrt{\left(\frac{v^2 \sin 	heta \cos 	heta}{g}ight)^2 + \left(\frac{v^2 \sin^2 	heta}{2g}ight)^2} = \frac{v^2 \sin 	heta}{g} \sqrt{\cos^2 	heta + \frac{\sin^2 	heta}{4}} = \frac{v^2 \sin 	heta}{2g} \sqrt{4 \cos^2 	heta + \sin^2 	heta} = \frac{v^2 \sin 	heta}{2g} \sqrt{3 \cos^2 	heta + 1}$.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 of $v$ at an angle of projection $\theta$. The average velocity of the particle between its time of projection and the time it reaches the highest point of its trajectory is

Similar Questions

$A$ projectile is thrown with an initial velocity of $10 \ m/s$ at an angle $\alpha$ with the horizontal. It has a range of $5 \ m$. Taking $g = 10 \ m/s^2$ and neglecting air resistance, what will be the estimated value of $\alpha$ (in $^{\circ}$)?

$A$ ball thrown by one player reaches the other in $2 \ s$. The maximum height attained by the ball above the point of projection will be about ....... $m$. (Take $g = 10 \ m/s^2$)

$A$ body is projected horizontally from a tower with a velocity of $18 \, ms^{-1}$. It strikes the ground at an angle of $45^o$. The vertical component of the velocity when it strikes the ground is ........ $ms^{-1}$.

$A$ projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then their product is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module