Two stones having different masses m_1 and m_ | vedclass

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

Question

(D) The maximum height $H$ attained by a projectile is given by the formula $H = \frac{u^2 \sin^2 	heta}{2g}$,where $u$ is the initial speed,$	heta$

Vedclass · Accepted Answer

$	an^2 \alpha: 1$

Vedclass · Answer

(D) The maximum height $H$ attained by a projectile is given by the formula $H = \frac{u^2 \sin^2 	heta}{2g}$,where $u$ is the initial speed,$	heta$ is the angle of projection,and $g$ is the acceleration due to gravity.For the first stone,the angle of projection is $\alpha$,so its maximum height is $H_1 = \frac{u^2 \sin^2 \alpha}{2g}$.For the second stone,the angle of projection is $(90^{\circ}-\alpha)$,so its maximum height is $H_2 = \frac{u^2 \sin^2(90^{\circ}-\alpha)}{2g}$.Since $\sin(90^{\circ}-\alpha) = \cos \alpha$,we have $H_2 = \frac{u^2 \cos^2 \alpha}{2g}$.The ratio of their maximum heights is $\frac{H_1}{H_2} = \frac{\frac{u^2 \sin^2 \alpha}{2g}}{\frac{u^2 \cos^2 \alpha}{2g}} = \frac{\sin^2 \alpha}{\cos^2 \alpha} = 	an^2 \alpha$.Thus,the ratio is $	an^2 \alpha : 1$.

Two stones having different masses $m_1$ and $m_2$ are projected at an angle $\alpha$ and $(90^{\circ}-\alpha)$ with the same speed from the same point. The ratio of their maximum heights is

Similar Questions

$A$ body is projected horizontally from the top of a tower with an initial velocity of $18 \, m s^{-1}$. It hits the ground at an angle of $45^{\circ}$. What is the vertical component of velocity when it strikes the ground? $(ms^{-1})$

$A$ cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

$A$ stone is projected in air. Its time of flight is $3\,s$ and range is $150\,m$. The maximum height reached by the stone is $......\,m$ $\left(g=10\,m/s^2\right)$.

At which points of its trajectory will a projectile have minimum and maximum speed?

$A$ projectile is projected at $10 \ m/s$ by making an angle of $60^{\circ}$ to the horizontal. After some time,its velocity makes an angle of $30^{\circ}$ to the horizontal. Its speed at this instant is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module