Two projectiles are fired with the same initi | vedclass

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

Question

(C) The formula for the maximum height attained by a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given the initial speed $u$ is the same for both

Vedclass · Accepted Answer

$\frac{1-\sin 2 \alpha}{1+\sin 2 \alpha}$

Vedclass · Answer

(C) The formula for the maximum height attained by a projectile is $H = \frac{u^2 \sin^2 	heta}{2g}$.Given the initial speed $u$ is the same for both projectiles,the ratio of their maximum heights is given by:$\frac{H_1}{H_2} = \frac{\sin^2(45^{\circ}-\alpha)}{\sin^2(45^{\circ}+\alpha)}$.Using the trigonometric expansion $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$:$\sin(45^{\circ}-\alpha) = \frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha = \frac{1}{\sqrt{2}}(\cos \alpha - \sin \alpha)$.$\sin(45^{\circ}+\alpha) = \frac{1}{\sqrt{2}}\cos \alpha + \frac{1}{\sqrt{2}}\sin \alpha = \frac{1}{\sqrt{2}}(\cos \alpha + \sin \alpha)$.Squaring these expressions:$\sin^2(45^{\circ}-\alpha) = \frac{1}{2}(\cos^2 \alpha + \sin^2 \alpha - 2 \sin \alpha \cos \alpha) = \frac{1}{2}(1 - \sin 2\alpha)$.$\sin^2(45^{\circ}+\alpha) = \frac{1}{2}(\cos^2 \alpha + \sin^2 \alpha + 2 \sin \alpha \cos \alpha) = \frac{1}{2}(1 + \sin 2\alpha)$.Therefore,the ratio is $\frac{H_1}{H_2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$.

Two projectiles are fired with the same initial speed from the same point on the ground at angles of $(45^{\circ}-\alpha)$ and $(45^{\circ}+\alpha)$,respectively,with the horizontal direction. The ratio of their maximum heights attained is:

Similar Questions

The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is

The equation of a projectile is $y = \sqrt{3} x - \frac{gx^2}{2}$. The angle of projection is:

If the kinetic energy of a second ball at its maximum height is $K$,what will be the kinetic energy of the first ball at its maximum height? (Assuming the first ball is thrown vertically upwards).

The angle of projection of a projectile whose path is shown in the given figure is

$A$ ball is projected with kinetic energy $E$ at an angle of $45^\circ$ to the horizontal. At the highest point during its flight,its kinetic energy will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module