### Related Formula For a projectile with launch speed u$u$ and angle theta$\theta$: - Horizontal Range: R = fracu^2 sin(2theta)g = frac2 u^2 sintheta costhetag$R = \frac{u^2 \sin(2\theta)}{g} = \frac{2 u^2 \sin\theta \cos\theta}{g}$ - Maximum Height: H = fracu^2 sin^2theta2g$H = \frac{u^2 \sin^2\theta}{2g}$ The general ratio linking range and maximum height is: tantheta = frac4HR$\tan\theta = \frac{4H}{R}$ ### Core Logic Given state: R = 3H Rightarrow fracHR = frac13$R = 3H \Rightarrow \frac{H}{R} = \frac{1}{3}$ ### Step 1: Determine the projection angle (theta$\theta$) Substitute the ratio into the relation: tantheta = 4 left(fracHRright) = 4 left(frac13right) = frac43$\tan\theta = 4 \left(\frac{H}{R}\right) = 4 \left(\frac{1}{3}\right) = \frac{4}{3}$ This is a standard Pythagorean triangle angle: sintheta = frac45, quad costheta = frac35$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}$ ### Step 2: Compute the horizontal range (R$R$) R = frac2 u^2 sintheta costhetag$R = \frac{2 u^2 \sin\theta \cos\theta}{g}$ R = frac2 u^2 left(frac45right) left(frac35right)g = frac24 u^225 g$R = \frac{2 u^2 \left(\frac{4}{5}\right) \left(\frac{3}{5}\right)}{g} = \frac{24 u^2}{25 g}$ Comparing this with the given format fracnu^225g$\frac{nu^2}{25g}$: n = 24$n = 24$ ### Pattern Recognition The relation tantheta = 4H/R$\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever R = k H$R = k H$, then tantheta = 4/k$\tan\theta = 4/k$. Recognizing standard angles like tantheta = 4/3$\tan\theta = 4/3$ or 3/4$3/4$ directly yields trigonometric values immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula The time of flight T$T$ of a projectile launched with speed u$u$ at an angle theta$\theta$ with the horizontal is: T = frac2u sinthetag$T = \frac{2u \sin\theta}{g}$ ### Core Logic The launch angles of the two projectiles are: $ ### Core Logic The launch angles of the two projectiles are: $theta_1 = 45^circ + alpha$\theta_1 = 45^\circ + \alpha$ $ $theta_2 = 45^circ - alpha$\theta_2 = 45^\circ - \alpha$ Since they have the same speed $ Since they have the same speed $u: $: $fracT_1T_2 = fracsin(45^circ + alpha)sin(45^circ - alpha)$\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$ ### Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas: $ ### Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas: $fracT_1T_2 = fracsin 45^circ cos alpha + cos 45^circ sin alphasin 45^circ cos alpha - cos 45^circ sin alpha$\frac{T_1}{T_2} = \frac{\sin 45^\circ \cos \alpha + \cos 45^\circ \sin \alpha}{\sin 45^\circ \cos \alpha - \cos 45^\circ \sin \alpha}$ $ $fracT_1T_2 = fracfrac1sqrt2cosalpha + frac1sqrt2sinalphafrac1sqrt2cosalpha - frac1sqrt2sinalpha = fraccosalpha + sinalphacosalpha - sinalpha$\frac{T_1}{T_2} = \frac{\frac{1}{\sqrt{2}}\cos\alpha + \frac{1}{\sqrt{2}}\sin\alpha}{\frac{1}{\sqrt{2}}\cos\alpha - \frac{1}{\sqrt{2}}\sin\alpha} = \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$ Divide numerator and denominator by $ Divide numerator and denominator by $\cos\alpha: $: $fracT_1T_2 = frac1 + tanalpha1 - tanalpha$\frac{T_1}{T_2} = \frac{1 + \tan\alpha}{1 - \tan\alpha}$$ ### Pattern Recognition Sees: Projectile angles complementary to $ ### Pattern Recognition Sees: Projectile angles complementary to $45^\circ. Shortcut: Remember the identity $. Shortcut: Remember the identity $\tan(45^\circ + \alpha) = \frac{1+\tan\alpha}{1-\tan\alpha}. Since complementary angles have sine ratios proportional to $. Since complementary angles have sine ratios proportional to $\sin(45^\circ + \alpha)/\sin(45^\circ - \alpha) = \tan(45^\circ + \alpha), the answer is directly $, the answer is directly $\frac{1+\tan\alpha}{1-\tan\alpha}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula x = u cdot t$x = u \cdot t$ [cite: 763] H = frac12gt^2$H = \frac{1}{2}gt^2$ [cite: 764] D = sqrtx^2 + H^2$D = \sqrt{x^2 + H^2}$ [cite: 774] ### Core Logic First, convert the horizontal velocity component to metric SI units: [cite: 157, 763] u = 360 times frac518 = 100\ textm/s$u = 360 \times \frac{5}{18} = 100\ \text{m/s}$ [cite: 157, 763] Calculate the horizontal range distance x$x$ covered over t = 20\ texts$t = 20\ \text{s}$: [cite: 158, 763] x = 100 times 20 = 2000\ textm = 2\ textkm$x = 100 \times 20 = 2000\ \text{m} = 2\ \text{km}$ [cite: 158, 763] The vertical displacement height H$H$ is explicitly given as 2\ textkm = 2000\ textm$2\ \text{km} = 2000\ \text{m}$[cite: 157, 769]. Let's confirm with free-fall height calculation matching the solution template: [cite: 764] H = frac12 times 10 times (20)^2 = 5 times 400 = 2000\ textm = 2\ textkm$H = \frac{1}{2} \times 10 \times (20)^2 = 5 \times 400 = 2000\ \text{m} = 2\ \text{km}$ [cite: 158, 769] Now find the net spatial vector displacement D$D$ from the release coordinates: [cite: 774] D = sqrtx^2 + H^2 = sqrt2^2 + 2^2 = sqrt8 = 2sqrt2\ textkm$D = \sqrt{x^2 + H^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}\ \text{km}$ [cite: 774] ### Pattern Recognition Be careful with the wording: the question asks for the displacement from the *release coordinate position* [cite: 159], which is the hypotenuse vector sqrtx^2 + H^2$\sqrt{x^2 + H^2}$[cite: 774]. Do not mistake it for the horizontal range distance alone. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula vecv = fracdvecrdt$\vec{v} = \frac{d\vec{r}}{dt}$ ### Core Logic Given position vector: vecr = 5t^2hati - 5thatj$\vec{r} = 5t^2\hat{i} - 5t\hat{j}$ Differentiating with respect to t$t$: vecv = 10thati - 5hatj$\vec{v} = 10t\hat{i} - 5\hat{j}$ At t = 2\ mathrms$t = 2\ \mathrm{s}$: vecv = 20hati - 5hatj$\vec{v} = 20\hat{i} - 5\hat{j}$ Magnitude of velocity: v = sqrt(20)^2 + (-5)^2 = sqrt400 + 25 = sqrt425 = 5sqrt17\ mathrmm/s$v = \sqrt{(20)^2 + (-5)^2} = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}\ \mathrm{m/s}$ Direction analysis:

v_x = 20$v_x = 20$ along +x$+x$ axis, v_y = -5$v_y = -5$ along -y$-y$ axis. Let theta$\theta$ be the angle made with the negative Y$Y$-axis: tan theta = left|fracv_xv_y ight| = frac205 = 4 implies theta = tan^-14$\tan \theta = \left|\frac{v_x}{v_y} ight| = \frac{20}{5} = 4 \implies \theta = \tan^{-1}4$ ### Pattern Recognition Always read the reference axis carefully in direction questions. Here, the angle is measured from the negative Y$Y$-axis, making tantheta = v_x / |v_y|$\tan\theta = v_x / |v_y|$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane