Two projectiles are fired with same initial speed from same point on ground at angles of (45^circ - alpha)$(45^{\circ} - \alpha)$ and (45^circ + alpha)$(45^{\circ} + \alpha)$ , respectively, with the horizontal direction. The ratio of their maximum heights attained is :

### Related Formula |vecA| = sqrtA_x^2 + A_y^2$$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$$ vecN = |vecA| hatB$$\vec{N} = |\vec{A}| \hat{B}$$ ### Core Logic We need to find a new vector vecN$\vec{N}$ that has the magnitude of vecA$\vec{A}$ and the direction of vecB$\vec{B}$. Magnitude of vecA$\vec{A}$: |vecA| = sqrt3^2 + 4^2 = sqrt25 = 5$|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Unit vector in the direction of vecB$\vec{B}$: hatB = fracvecB|vecB| = frac4hati + 3hatjsqrt4^2 + 3^2 = frac4hati + 3hatj5$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{\sqrt{4^2 + 3^2}} = \frac{4\hat{i} + 3\hat{j}}{5}$. ### Step 1: Construct the Vector vecN = |vecA| hatB = 5 left( frac4hati + 3hatj5 right)$$\vec{N} = |\vec{A}| \hat{B} = 5 \left( \frac{4\hat{i} + 3\hat{j}}{5} \right)$$ vecN = 4hati + 3hatj$$\vec{N} = 4\hat{i} + 3\hat{j}$$ ### Step 2: Match Components The x$x$ and y$y$ components are given as x$x$ and 3$3$. From vecN = 4hati + 3hatj$\vec{N} = 4\hat{i} + 3\hat{j}$, we see the x$x$-component is 4$4$. Therefore, x = 4$x = 4$. ### Pattern Recognition Constructing a vector matching magnitude and direction is a simple scalar multiplication of the desired magnitude by the target direction's unit vector. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula The magnitude of the resultant vector is given by: |vecR_res| = sqrtA^2 + B^2 + 2AB cos theta$$|\vec{R}_{res}| = \sqrt{A^2 + B^2 + 2AB \cos \theta}$$ ### Core Logic Let |vecA| = |vecB| = R$|\vec{A}| = |\vec{B}| = R$. Then for vector addition: |vecA + vecB| = sqrtR^2 + R^2 + 2R^2 cos theta$$|\vec{A} + \vec{B}| = \sqrt{R^2 + R^2 + 2R^2 \cos \theta}$$ ### Step 1: Simplify Addition Form |vecA + vecB| = sqrt2R^2 (1 + cos theta)$$|\vec{A} + \vec{B}| = \sqrt{2R^2 (1 + \cos \theta)}$$ Using the trigonometric identity 1 + cos theta = 2 cos^2 left(fractheta2right)$1 + \cos \theta = 2 \cos^2 \left(\frac{\theta}{2}\right)$: |vecA + vecB| = sqrt2R^2 times 2 cos^2 left(fractheta2right) = 2R cos left(fractheta2right)$$|\vec{A} + \vec{B}| = \sqrt{2R^2 \times 2 \cos^2 \left(\frac{\theta}{2}\right)} = 2R \cos \left(\frac{\theta}{2}\right)$$ ### Step 2: Cross-check Subtraction Form For subtraction: |vecA - vecB| = sqrtR^2 + R^2 - 2R^2 cos theta$$|\vec{A} - \vec{B}| = \sqrt{R^2 + R^2 - 2R^2 \cos \theta}$$ |vecA - vecB| = sqrt2R^2 (1 - cos theta) = sqrt2R^2 times 2 sin^2 left(fractheta2right) = 2R sin left(fractheta2right)$$|\vec{A} - \vec{B}| = \sqrt{2R^2 (1 - \cos \theta)} = \sqrt{2R^2 \times 2 \sin^2 \left(\frac{\theta}{2}\right)} = 2R \sin \left(\frac{\theta}{2}\right)$$ Checking options, only |vecA + vecB| = 2R cos left(fractheta2right)$|\vec{A} + \vec{B}| = 2R \cos \left(\frac{\theta}{2}\right)$ is correctly paired in the choice list. ### Pattern Recognition Standard geometry shortcut: Addition of two equal vectors yields a cosine half-angle dependency. Subtraction yields a sine half-angle dependency. (+ to cos)$(+ \to \cos)$, (- to sin)$(- \to \sin)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane