A helicopter flying horizontally with a speed of 360~km/h$360~km/h$ at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is: (use acceleration due to gravity g=10~m/s^2$g=10~m/s^{2}$ and neglect air resistance) [cite: 157, 158, 159, 160]

### Related Formula T = frac2u_yg, quad R = v_x cdot T$$T = \frac{2u_y}{g}, \quad R = v_x \cdot T$$ ### Core Logic

The total horizontal velocity components combine due to downstream motion addition v_x = 27 + 9 = 36text km/h = 36 cdot frac518 = 10text m/s$$v_x = 27 + 9 = 36\text{ km/h} = 36 \cdot \frac{5}{18} = 10\text{ m/s}$$ The time of flight for the vertical launch tracking stands at : T = frac2 cdot 1010 = 2text s$$T = \frac{2 \cdot 10}{10} = 2\text{ s}$$ Horizontal range measured along the river bank frame is : Range = v_x cdot T = 10 cdot 2 = 20text m = 2000text cm$$Range = v_x \cdot T = 10 \cdot 2 = 20\text{ m} = 2000\text{ cm}$$ ### Chapter Mix Class 11 Physics: Motion in a Straight Line 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