If the maximum height and range of a projecti | vedclass

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

Question

(C) Given: Maximum height $H = 3 \,m$,Range $R = 4 \,m$,and $g = 10 \,ms^{-2}$.We know that $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin

Vedclass · Accepted Answer

$10 \sqrt{\frac{2}{3}} \,ms^{-1}$

Vedclass · Answer

(C) Given: Maximum height $H = 3 \,m$,Range $R = 4 \,m$,and $g = 10 \,ms^{-2}$.We know that $H = \frac{u^2 \sin^2 	heta}{2g}$ and $R = \frac{u^2 \sin 2	heta}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.Taking the ratio of $H$ and $R$:$\frac{H}{R} = \frac{u^2 \sin^2 	heta / 2g}{2u^2 \sin 	heta \cos 	heta / g} = \frac{\sin 	heta}{4 \cos 	heta} = \frac{1}{4} 	an 	heta$.Substituting the values: $\frac{3}{4} = \frac{1}{4} 	an 	heta \Rightarrow 	an 	heta = 3$.Since $	an 	heta = 3$,we have $\sin 	heta = \frac{3}{\sqrt{10}}$ and $\cos 	heta = \frac{1}{\sqrt{10}}$.Using the formula for $H$:$3 = \frac{u^2 \sin^2 	heta}{2g} = \frac{u^2 (3/\sqrt{10})^2}{2 	imes 10} = \frac{u^2 	imes (9/10)}{20}$.$3 = \frac{9u^2}{200} \Rightarrow u^2 = \frac{3 	imes 200}{9} = \frac{200}{3}$.$u = \sqrt{\frac{200}{3}} = 10 \sqrt{\frac{2}{3}} \,ms^{-1}$.Thus,the correct option is $C$.

If the maximum height and range of a projectile are $3 \,m$ and $4 \,m$ respectively,then the velocity of the projectile is (Take $g=10 \,ms^{-2}$)

Similar Questions

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

The position of a projectile launched from the origin at $t = 0$ is given by $\vec{r} = (40\hat{i} + 50\hat{j})\,m$ at $t = 2\,s$. If the projectile was launched at an angle $\theta$ from the horizontal,then $\theta$ is (take $g = 10\,m/s^2$)

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The maximum height of the projectile is ........ $m$.

The maximum height reached by a projectile is $64 \,m$. If the initial velocity is halved, the new maximum height of the projectile is . . . . . . $m$.

$A$ particle is projected with speed $u$ at an angle $\theta$ with the horizontal from the ground. If it is at the same height from the ground at times $t_1$ and $t_2$,then its average velocity in the time interval $t_1$ to $t_2$ is .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module