The path of a projectile is given by the equa | vedclass

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

Question

(D) The given equation of the trajectory is $y = ax - bx^2$.The standard equation of a projectile trajectory is $y = x 	an 	heta - \frac{gx^2}{2u^2

Vedclass · Accepted Answer

$\frac{a^2}{4b}, 	an^{-1}(a)$

Vedclass · Answer

(D) The given equation of the trajectory is $y = ax - bx^2$.The standard equation of a projectile trajectory is $y = x 	an 	heta - \frac{gx^2}{2u^2 \cos^2 	heta}$.Comparing the coefficients of $x$ and $x^2$ in both equations:$	an 	heta = a \implies 	heta = 	an^{-1}(a)$.Also, $b = \frac{g}{2u^2 \cos^2 	heta}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.From $	an 	heta = a$, we have $\sin 	heta = \frac{a}{\sqrt{1+a^2}}$ and $\cos 	heta = \frac{1}{\sqrt{1+a^2}}$.Substituting $u^2 = \frac{g}{2b \cos^2 	heta}$ into the height formula:$H = \frac{g}{2b \cos^2 	heta} \cdot \frac{\sin^2 	heta}{2g} = \frac{	an^2 	heta}{4b} = \frac{a^2}{4b}$.Thus, the maximum height is $\frac{a^2}{4b}$ and the angle of projection is $	an^{-1}(a)$.

Similar Questions

$A$ particle is projected from a horizontal plane such that its velocity vector at time $t$ is given by $\vec{v} = a\hat{i} + (b - ct)\hat{j}$. Its range on the horizontal plane is given by

Two projectiles,one fired from the surface of the Earth with velocity $10 \, m/s$ and the other fired from the surface of another planet with initial speed $5 \, m/s$,trace identical trajectories. The value of the acceleration due to gravity on the planet is ......... $m/s^2$.

$A$ projectile is thrown into space so as to have a maximum possible horizontal range of $400 \, m$. Taking the point of projection as the origin,the coordinates of the point where the velocity of the projectile is minimum are

Two projectiles are projected with the same initial velocities at $15^{\circ}$ and $30^{\circ}$ with respect to the horizontal. The ratio of their ranges is $1 : x$. The value of $x$ is:

$A$ particle is projected with velocity $u$ at an angle $\theta$ with the horizontal. What is the change in its velocity at the maximum height?

Vedclass Test Series

Exam Paper Generator

Online Exam Module