The Question is from neet-physics atomic-structure horizontal-projectile, and it was created by the author P.NARASIMHA

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

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: C
## CONCEPT FRAMEWORK
	**Primary Concept**: Projectile Motion
	**Related Topics**: Kinematics, Trigonometry, Constant Acceleration
	**Prerequisites**: Understanding of projectile motion equations, trigonometric functions, and basic kinematics.
## SOLUTION PATHWAY
**Given Information**:
	Initial velocity: $v_0$
	Launch angle: $	heta$
	Horizontal distance to cliff: D
	Acceleration due to gravity: g
**Method & Steps**:
	1.  Find the time it takes for the shell to travel a horizontal distance D.
		Reasoning: Horizontal motion is uniform, so we use the horizontal velocity component to find time.
		Formula/Rule: $x = v_{0x}t$, where $v_{0x} = v_0 \cos 	heta$
		Working: $D = v_0 \cos 	heta * t$, therefore $t = \frac{D}{v_0 \cos 	heta}$
	2. Find the vertical position of the shell at that time.
		Reasoning: Vertical motion is affected by gravity, so we 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: $y = v_0 \sin 	heta * \frac{D}{v_0 \cos 	heta} - \frac{1}{2}g(\frac{D}{v_0 \cos 	heta})^2$
		Key Point: Substituting the value of t calculated in step 1.
	3.  Simplify the equation.
		Working: $y = D 	an 	heta - \frac{1}{2}\frac{gD^2}{v_0^2 \cos^2 	heta}$
		Result: The height at which the cannon strikes the cliff is $D	an 	heta - \frac{1}{2}\frac{g{D^2}}{v_0^2{{\cos }^2}	heta }$.
		Verification: The derived equation matches option C.
 ## OPTION ANALYSIS
[A] :x: Incorrect. The first term should be $D	an	heta$, not $D\sin	heta$, and the denominator of the second term should have $\cos^2	heta$ not $\sin^2	heta$.
[B] :x: Incorrect. The first term should be $D	an	heta$, not $D\cos	heta$.
[C] ✓: Correct. This matches our derived equation.
[D] :x: Incorrect. The denominator of the second term should have $\cos^2	heta$ not $\sin^2	heta$.
 ## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Using sine for horizontal velocity and cosine for vertical velocity. [How to avoid] Remember that horizontal velocity is always associated with cosine ($v_{0x} = v_0\cos	heta$) and vertical velocity with sine ($v_{0y} = v_0\sin	heta$).
		2.[Error 2]: Forgetting the negative sign in front of the gravity term. [How to avoid]  The acceleration due to gravity acts downwards, opposing the initial vertical velocity, so it must be negative in the vertical displacement equation.
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the time of flight equation in terms of horizontal distance: $t = \frac{D}{v_0\cos	heta}$.
		2.[Memory aid]: The vertical motion equation involves $v_{0y}t - \frac{1}{2}gt^2$ .Substitute the time equation to obtain the correct answer.

P.NARASIMHA · Answer

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

P.NARASIMHA · Answer

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

P.NARASIMHA · Answer

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

P.NARASIMHA · Answer

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

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