The Question is from neet-physics motion-in-a-plane , and it was created by the author NAGA SAI

Question

A cannon on a level plane is aimed at an angle $	heta $ above the horizontal and a shell is fired with muzzle velocity ${v_0}$ towards a cliff D distance away. The height at which the cannon strikes the cliff is given by

NAGA SAI · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Projectile motion under constant gravitational acceleration.
	**Related Topics**: Kinematic equations, resolving initial velocity into horizontal and vertical components, time of flight.
	**Prerequisites**: Basic knowledge of projectile motion, trigonometric functions, and kinematic equations for constant acceleration.
## SOLUTION PATHWAY
**Given Information**:
	* Initial velocity: $v_0$
	* Angle of projection: $	heta$
	* Horizontal distance to cliff: D
	* Acceleration due to gravity: g
**Method & Steps**:
	1. **Horizontal Motion:**
		Reasoning: The horizontal velocity is constant. We can find the time it takes to reach the cliff using the horizontal distance and velocity.
		Formula/Rule: $x = v_{0x} t$ where $v_{0x} = v_0 \cos(	heta)$
		Working: $D = v_0 \cos(	heta) t  => t = \frac{D}{v_0 \cos(	heta)}$
	2. **Vertical Motion:**
		Reasoning: The vertical motion is influenced by gravity, and we can use the kinematic equation for vertical displacement.
		Formula/Rule: $y = v_{0y}t - \frac{1}{2}gt^2$ where $v_{0y} = v_0 \sin(	heta)$
		Working: Substituting t from step 1:
		$y = v_0 \sin(	heta) \frac{D}{v_0 \cos(	heta)} - \frac{1}{2}g(\frac{D}{v_0 \cos(	heta)})^2$
		Key Point: Using the trigonometric identity $ \frac{\sin(	heta)}{\cos(	heta)} = 	an(	heta)$ simplifies the first term.
	3. **Final Step:**
		Result: $y = D	an(	heta) - \frac{1}{2}\frac{gD^2}{v_0^2\cos^2(	heta)}$
		Verification: The derived equation matches option C.
 ## OPTION ANALYSIS
[A] :x: Incorrect due to using sine for horizontal distance and incorrect time component in the vertical motion part.
[B] :x: Incorrect due to using cosine for horizontal distance and incorrect time component in the vertical motion part.
[C] ✓: Correct as it matches the derived equation.
[D] :x: Incorrect because it uses sine instead of cosine in the denominator of the acceleration term.
 ## LEARNING AIDS
	**Common Mistakes**:
		1. **Incorrect trigonometric components:** Using sine for horizontal and cosine for vertical components.
			How to avoid: Remember that cosine is associated with the adjacent side (horizontal) and sine with the opposite side (vertical).
		2. **Incorrectly combining horizontal and vertical motion**: Forgetting that the time to travel horizontally is the same as the time to travel vertically.
			How to avoid: Use the time found from the horizontal motion in the vertical motion equation.
	**Quick Tips**:
		1. Break down projectile motion into horizontal and vertical components.
		2. Use the correct kinematic equations with initial velocities and acceleration components.

NAGA SAI · Answer

$D\sin 	heta - \frac{1}{2}\frac{{g{D^2}}}{{v_0^2{{\sin }^2}	heta }}$

NAGA SAI · Answer

$D\cos 	heta - \frac{1}{2}\frac{{g{D^2}}}{{v_0^2{{\cos }^2}	heta }}$

NAGA SAI · Answer

The correct Option is "$D	an 	heta - \frac{1}{2}\frac{{g{D^2}}}{{v_0^2{{\cos }^2}	heta }}$"

NAGA SAI · Answer

$D	an 	heta - \frac{1}{2}\frac{{g{D^2}}}{{v_0^2{{\sin }^2}	heta }}$

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