Question 17: Motion in 2D & Vectors

Question

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^2}{25 g}$, where value of $n$ is: (Given, ' $g$ ' is the acceleration due to gravity.)

Accepted Answer

Given that the horizontal range of a projectile is three times its maximum height, we need to determine the value of $ n $ in the expression for the range $\frac{n u^2}{25 g}$.
Let's start by setting up the equations for the range and maximum height of a projectile.
The formula for the range $ R $ is:
$ R = \frac{u^2 \sin 2	heta}{g} $
The formula for the maximum height $ H_{\max} $ is:
$ H_{\max} = \frac{u^2 \sin^2 	heta}{2g} $
According to the problem, the range $ R $ is three times the maximum height $ H_{\max} $:
$ R = 3H_{\max} $
Substituting the formulas for $ R $ and $ H_{\max} $ into this condition gives:
$ \frac{u^2 \sin 2	heta}{g} = 3 \cdot \frac{u^2 \sin^2 	heta}{2g} $
Simplifying this equation, we get:
$ 2 \sin 	heta \cos 	heta = \frac{3}{2} \sin^2 	heta $
Using the identity $ \sin 2	heta = 2 \sin 	heta \cos 	heta $, we can rewrite:
$ 2 \sin 	heta \cos 	heta = \sin 2	heta $
This gives us:
$ \sin 2	heta = \frac{3}{2} \sin^2 	heta $
Dividing both sides by $\sin 	heta$, assuming $\sin 	heta 
eq 0$, we have:
$ 2 \cos 	heta = \frac{3}{2} \sin 	heta $
Rearranging the terms to solve for $	an 	heta$, we get:
$ 	an 	heta = \frac{4}{3} $
Thus, the angle $	heta$ where $	an 	heta = \frac{4}{3}$ corresponds to $	heta = 53^\circ$.
Now, substituting back into the range formula:
$ R = \frac{u^2 \cdot 2 \cdot \frac{3}{5} \cdot \frac{4}{5}}{g} $
The calculation yields:
$ R = \frac{24 u^2}{25 g} $
Therefore, the value of $ n $ in the expression $\frac{n u^2}{25 g}$ is $ n = 24 $.

Answer

6

Answer

12

Answer

18

Answer

24

Motion in 2D & Vectors

A particle is projected with velocity u so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as \frac{n u^2}{25 g}, where value of n is: (Given, ' g ' is the acceleration due to gravity.)

Choose the correct answer: