The Question is from neet-physics motion-in-a-straight-line , and it was created by the author ABDUL JILANI BASHA

Question

A projectile is launched at an angle $	heta$ with the horizontal with an initial speed $u$. If the horizontal range is $R$ and the maximum height reached is $H$, what is the relationship between $R$, $H$, and $	heta$?

ABDUL JILANI BASHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [Option C]
## CONCEPT FRAMEWORK
	**Primary Concept**: Projectile Motion
	**Related Topics**: Kinematics, Trigonometry
	**Prerequisites**: Understanding of projectile motion equations
## SOLUTION PATHWAY
**Given Information**:
	* Initial speed: $u$
	* Launch angle: $	heta$
	* Horizontal range: $R$
	* Maximum height: $H$

**Method & Steps**:
	1. Express the horizontal range $R$ in terms of $u$ and $	heta$.
		Reasoning: The horizontal range is the horizontal distance covered by the projectile when it returns to the same vertical level it was launched from.
		Formula/Rule: $R = \frac{u^2 \sin(2	heta)}{g}$
	2. Express the maximum height $H$ in terms of $u$ and $	heta$.
		Reasoning: The maximum height is the vertical distance the projectile reaches at the highest point of its trajectory.
		Formula/Rule: $H = \frac{u^2 \sin^2(	heta)}{2g}$
	3. Find a relationship between $R$ and $H$.
		Working: Divide the equation for $R$ by the equation for $H$.
			$\frac{R}{H} = \frac{\frac{u^2 \sin(2	heta)}{g}}{\frac{u^2 \sin^2(	heta)}{2g}} = \frac{2\sin(2	heta)}{\sin^2(	heta)} = \frac{2(2\sin(	heta)\cos(	heta))}{\sin^2(	heta)} = \frac{4\cos(	heta)}{\sin(	heta)} = 4\cot(	heta)$
	4. Express $R$ in terms of $H$ and $	heta$.
		Result: $R = 4H \cot(	heta)$
		Verification: The derived relation matches the correct option.
## OPTION ANALYSIS
[A] :x: The relation derived is $R = 4H \cot(	heta)$, not $R = 2H 	an(	heta)$.
[B] :x: The relation derived is $R = 4H \cot(	heta)$, not $R = H \cot(	heta)$.
[C] ✓: The relation derived $R = 4H \cot(	heta)$ matches this option.
[D] :x: The relation derived is $R = 4H \cot(	heta)$, not $R = 4H 	an(	heta)$.
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Incorrectly using the formulas for range and height. [How to avoid]: Memorize or correctly derive the range and height equations.
		2.[Error 2]: Making errors in trigonometric simplifications. [How to avoid]: Practice trigonometric identities.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the relationship between $\sin(2	heta)$ and $\sin(	heta)\cos(	heta)$.
		2.[Memory aid]: Visualize the projectile motion and its maximum height and range to aid understanding.

ABDUL JILANI BASHA · Answer

$R = 2H 	an 	heta$

ABDUL JILANI BASHA · Answer

$R = H \cot 	heta$

ABDUL JILANI BASHA · Answer

The correct Option is "$R = 4H \cot 	heta$"

ABDUL JILANI BASHA · Answer

$R = 4H 	an 	heta$

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