### Related Formula H_max = fracu^2 sin^2 theta2g$H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$ ### Core Logic Let theta_1 = 45^circ - alpha$\theta_1 = 45^{\circ} - \alpha$ and theta_2 = 45^circ + alpha$\theta_2 = 45^{\circ} + \alpha$. The ratio of maximum heights is: fracH_1H_2 = fracsin^2(45^circ - alpha)sin^2(45^circ + alpha)$\frac{H_1}{H_2} = \frac{\sin^2(45^{\circ} - \alpha)}{\sin^2(45^{\circ} + \alpha)}$ ### Step 1: Simplify Trigonometric Terms sin(45^circ pm alpha) = frac1sqrt2cosalpha pm frac1sqrt2sinalpha$\sin(45^{\circ} \pm \alpha) = \frac{1}{\sqrt{2}}\cos\alpha \pm \frac{1}{\sqrt{2}}\sin\alpha$ fracH_1H_2 = frac(cosalpha - sinalpha)^2(cosalpha + sinalpha)^2 = fraccos^2alpha + sin^2alpha - 2sinalphacosalphacos^2alpha + sin^2alpha + 2sinalphacosalpha = frac1 - sin 2alpha1 + sin 2alpha$\frac{H_1}{H_2} = \frac{(\cos\alpha - \sin\alpha)^2}{(\cos\alpha + \sin\alpha)^2} = \frac{\cos^2\alpha + \sin^2\alpha - 2\sin\alpha\cos\alpha}{\cos^2\alpha + \sin^2\alpha + 2\sin\alpha\cos\alpha} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$ [cite: 602, 603] ### Pattern Recognition For complementary angles shifted symmetric to 45^circ$45^{\circ}$, ratios simplify cleanly via sin 2alpha$\sin 2\alpha$ identities[cite: 602, 603]. ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula Uniform circular velocity: v = frac2pi RT$v = \frac{2\pi R}{T}$ Maximum height of a projectile: H = fracv^2 sin^2theta2g$H = \frac{v^2 \sin^2\theta}{2g}$ ### Core Logic Given that the projectile max height matches H = 4R$H = 4R$: 4R = fracv^2 sin^2theta2g$4R = \frac{v^2 \sin^2\theta}{2g}$ Substitute the value of v$v$ from the circular motion loop: 4R = fracleft(frac2pi RTright)^2 sin^2theta2g$4R = \frac{\left(\frac{2\pi R}{T}\right)^2 \sin^2\theta}{2g}$ 4R = frac4pi^2 R^2 sin^2theta2g T^2$4R = \frac{4\pi^2 R^2 \sin^2\theta}{2g T^2}$ ### Step 1: Isolate Angular Components Cancel out 4R$4R$ from both sides: 1 = fracpi^2 R sin^2theta2g T^2$1 = \frac{\pi^2 R \sin^2\theta}{2g T^2}$ sin^2theta = frac2g T^2pi^2 R$\sin^2\theta = \frac{2g T^2}{\pi^2 R}$ sintheta = left(frac2g T^2pi^2 Rright)^frac12 implies theta = sin^-1left[frac2gT^2pi^2 Rright]^frac12$\sin\theta = \left(\frac{2g T^2}{\pi^2 R}\right)^{\frac{1}{2}} \implies \theta = \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{\frac{1}{2}}$ ### Pattern Recognition Connect circular metrics directly to projectile parameters via velocity matching. Keeping terms unsimplified makes it easy to cancel common elements later. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula The centripetal force (F$F$) acting on a particle of mass m$m$ moving with velocity v$v$ in a path of radius r$r$ is given by: F = fracm v^2r$F = \frac{m v^2}{r}$ ### Core Logic Given that the two particles have the same mass (m_1 = m_2$m_1 = m_2$) and the ratio of their radii of curvature is: fracr_1r_2 = frac34$\frac{r_1}{r_2} = \frac{3}{4}$ To maintain a constant (equal) centripetal force (F_1 = F_2$F_1 = F_2$): ### Step 1: Relate Velocity to Radius fracm_1 v_1^2r_1 = fracm_2 v_2^2r_2$\frac{m_1 v_1^2}{r_1} = \frac{m_2 v_2^2}{r_2}$ Since m_1 = m_2$m_1 = m_2$, the equation simplifies to: fracv_1^2r_1 = fracv_2^2r_2 implies fracv_1^2v_2^2 = fracr_1r_2$\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2} \implies \frac{v_1^2}{v_2^2} = \frac{r_1}{r_2}$ fracv_1v_2 = sqrtfracr_1r_2$\frac{v_1}{v_2} = \sqrt{\frac{r_1}{r_2}}$ ### Step 2: Calculate the Ratio Substituting the given ratio of radii: fracv_1v_2 = sqrtfrac34 = fracsqrt32$\frac{v_1}{v_2} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ Therefore, the ratio of their velocities is sqrt3:2$\sqrt{3}:2$. ### Pattern Recognition For problems involving steady values under constraint variations, establish the proportionality relation first. Here, F propto fracv^2r implies v propto sqrtr$F \propto \frac{v^2}{r} \implies v \propto \sqrt{r}$ when F$F$ and m$m$ are held constant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Related Formula For a projectile fired horizontally from height h$h$ with speed u$u$: textHorizontal Range R = u cdot t$\text{Horizontal Range } R = u \cdot t$ textVertical Displacement h = frac12 g t^2$\text{Vertical Displacement } h = \frac{1}{2} g t^2$ ### Core Logic To just clear step 4 and land on the tread of step 5, the flight trajectory must pass beyond the outer corner boundary vertex of the 4^textth$4^{\text{th}}$ step. Therefore, net horizontal distance required to cross 4 complete steps is: R = 4 times 0.1 mathrm~m = 0.4 mathrm~m$R = 4 \times 0.1 \mathrm{~m} = 0.4 \mathrm{~m}$ Similarly, net vertical fall matching the top of the 4^textth$4^{\text{th}}$ step level is: h = 4 times 0.1 mathrm~m = 0.4 mathrm~m$h = 4 \times 0.1 \mathrm{~m} = 0.4 \mathrm{~m}$

### Step 1: Determine Time of Flight Using the vertical kinematic equation: 0.4 = frac12 times 10 times t^2 implies 0.4 = 5 t^2$0.4 = \frac{1}{2} \times 10 \times t^2 \implies 0.4 = 5 t^2$ t^2 = frac0.45 = 0.08 mathrm~s^2$t^2 = \frac{0.4}{5} = 0.08 \mathrm{~s^2}$ ### Step 2: Determine Velocity Using the horizontal path equation: R = u cdot t implies R^2 = u^2 cdot t^2$R = u \cdot t \implies R^2 = u^2 \cdot t^2$ Substituting range and time values: (0.4)^2 = u^2 times 0.08$(0.4)^2 = u^2 \times 0.08$ 0.16 = u^2 times 0.08$0.16 = u^2 \times 0.08$ u^2 = frac0.160.08 = 2 implies u = sqrt2 mathrm~m/s$u^2 = \frac{0.16}{0.08} = 2 \implies u = \sqrt{2} \mathrm{~m/s}$ ### Step 3: Extract x Matching the parameter form u = sqrtx$u = \sqrt{x}$, we find: x = 2$x = 2$ ### Pattern Recognition To clear the n$n$-th step, the projectile must safely pass the outer point of step (n-1)$(n-1)$. Treat the geometric coordinates of corners as bounding constraints (x = (n-1)w$x = (n-1)w$, y = (n-1)h$y = (n-1)h$) to configure kinematics instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane

### Core Logic

For two projectiles launched from the same height to have the same time of flight (T$T$), their vertical components of velocity must be equal. Since the angles given are with the *vertical*, the vertical components are v_A cos(45^circ)$v_A \cos(45^{\circ})$ and v_B cos(60^circ)$v_B \cos(60^{\circ})$. If T_A = T_B$T_A = T_B$, then v_Ay = v_By$v_{Ay} = v_{By}$. v_A cos(45^circ) = v_B cos(60^circ)$v_A \cos(45^{\circ}) = v_B \cos(60^{\circ})$ v_A left(frac1sqrt2right) = v_B left(frac12right)$v_A \left(\frac{1}{\sqrt{2}}\right) = v_B \left(\frac{1}{2}\right)$ fracv_Av_B = fracsqrt22 = frac1sqrt2$\frac{v_A}{v_B} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$ ### Step 1: Check Inconsistency For the ranges to be the same while having the same time of flight, their horizontal components of velocity must also be equal: v_Ax = v_Bx$v_{Ax} = v_{Bx}$. v_A sin(45^circ) = v_B sin(60^circ)$v_A \sin(45^{\circ}) = v_B \sin(60^{\circ})$ v_A left(frac1sqrt2right) = v_B left(fracsqrt32right)$v_A \left(\frac{1}{\sqrt{2}}\right) = v_B \left(\frac{\sqrt{3}}{2}\right)$ fracv_Av_B = sqrtfrac32$\frac{v_A}{v_B} = \sqrt{\frac{3}{2}}$ This yields a contradiction. It is impossible for both the ranges and the times of flight to be simultaneously equal for different angles of projection from a tower. ### Step 2: Conclusion The question contains inconsistent data and is technically a Bonus question. However, if one arbitrarily equates only the time of flight (or if NTA intended a different scenario), the ratio fracv_Av_B = frac1sqrt2$\frac{v_A}{v_B} = \frac{1}{\sqrt{2}}$ matches option (4), which was the officially provided key before corrections. ### Pattern Recognition Be wary of over-constrained physics problems. If a question specifies both range AND time of flight are identical for two different angles, check if the math yields a contradiction. NTA often accepts the result of one partial constraint. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane