A stone projected with a velocity u at an ang | vedclass

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

Question

(A) The maximum height reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first case,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.

Vedclass · Accepted Answer

$R = 4\sqrt{H_1 H_2}$

Vedclass · Answer

(A) The maximum height reached by a projectile is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.For the first case,$H_1 = \frac{u^2 \sin^2 	heta}{2g}$.For the second case,the angle is $(\frac{\pi}{2} - 	heta)$,so $H_2 = \frac{u^2 \sin^2(\frac{\pi}{2} - 	heta)}{2g} = \frac{u^2 \cos^2 	heta}{2g}$.Multiplying $H_1$ and $H_2$,we get $H_1 H_2 = \frac{u^4 \sin^2 	heta \cos^2 	heta}{4g^2} = \frac{u^4 (2 \sin 	heta \cos 	heta)^2}{16g^2} = \frac{(u^2 \sin 2	heta)^2}{16g^2}$.Since the horizontal range $R = \frac{u^2 \sin 2	heta}{g}$,we have $H_1 H_2 = \frac{R^2}{16}$.Therefore,$R^2 = 16 H_1 H_2$,which implies $R = 4\sqrt{H_1 H_2}$.

Similar Questions

Two balls are thrown as shown in the figure and they reach the ground in the same time. What is the ratio of the maximum heights attained by them?

$A$ car is moving with a velocity of $4 \,m \,s^{-1}$ towards east. After a time of $4 \,s$, if it is heading north-east with a velocity of $4 \sqrt{2} \,m \,s^{-1}$, then the average velocity of the car is

Two particles move on a circular path (one just inside and the other just outside) with angular velocities $\omega$ and $5\omega$ starting from the same point. Then

$A$ particle is moving in a circle of radius $R$ with constant speed $V$. The magnitude of average acceleration after half revolution is

Vedclass Test Series

Exam Paper Generator

Online Exam Module