The horizontal range is four times the maximu | vedclass

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

Question

(C) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2	heta)}{g} = \frac{2u^2 \sin 	heta \cos 	heta}{g}$.The maximum height $H$ is given by $H = \frac{u^2 \sin^2 	heta}{2g}$.Given that the horizontal range is four times the maximum height,we have $R = 4H$.Substituting the formulas: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = 4 \left( \frac{u^2 \sin^2 	heta}{2g} ight)$.Simplifying the equation: $\frac{2u^2 \sin 	heta \cos 	heta}{g} = \frac{2u^2 \sin^2 	heta}{g}$.This leads to $\cos 	heta = \sin 	heta$,which implies $	an 	heta = 1$.Therefore,$	heta = 45^\circ$.

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is ......... $^o$

Similar Questions

Which one of the following is the most widely accepted method of contraception in India,at present?

Which one of the following statements is not correct?

In the coordination compound $K_4[Ni(CN)_4]$,the oxidation state of nickel is:

Which of the following sets of variables gives a straight line with a negative slope when plotted? ($P =$ vapour pressure,$T =$ Temperature in $K$)

What is the alum used for the purification of water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module