A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 \, mathrmm/s^2$3 \, \mathrm{m/s}^{2}$ . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s. The initial height of the aeroplane is ____ m. (g = 10 m/s ^2$^{2}$ )

### Related Formula t_textmin = fracdv_br$$t_{\text{min}} = \frac{d}{v_{br}}$$ textDrift (displacement along bank) = v_r times t_texttotal$$\text{Drift (displacement along bank)} = v_r \times t_{\text{total}}$$ ### Core Logic First, convert speeds to m/s. River speed v_R = 18 times frac518 = 5 text m/s$v_R = 18 \times \frac{5}{18} = 5 \text{ m/s}$. Boat speed in still water v_BR = 36 times frac518 = 10 text m/s$v_{BR} = 36 \times \frac{5}{18} = 10 \text{ m/s}$. River width d = 200 text m$d = 200 \text{ m}$.

For minimum time across the river, the boat must head exactly perpendicular to the river flow. ### Step 1: Calculating Minimum Time Time taken for one crossing (bank to bank): t_textone-way = fracdv_BR = frac20010 = 20 text s$$t_{\text{one-way}} = \frac{d}{v_{BR}} = \frac{200}{10} = 20 \text{ s}$$ Since it is a round trip (bank to bank and back), total minimum time is: t_texttotal = 2 times 20 = 40 text s$$t_{\text{total}} = 2 \times 20 = 40 \text{ s}$$ ### Step 2: Calculating Displacement along Bank Displacement along the river bank (drift) happens exclusively due to the river's flow during this total time: textDisplacement = v_R times t_texttotal = 5 times 40 = 200 text m$$\text{Displacement} = v_R \times t_{\text{total}} = 5 \times 40 = 200 \text{ m}$$ ### Pattern Recognition Minimum time always ignores drift and maxes out the perpendicular velocity vector. Drift simply becomes V_textriver times T_texttotal$V_{\text{river}} \times T_{\text{total}}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula v_perp = v_br sintheta$$v_{\perp} = v_{br} \sin\theta$$ w = v_perp cdot t$$w = v_{\perp} \cdot t$$ ### Core Logic Let the West-to-East river flow direction be along the positive x$x$-axis. The boat's velocity relative to water is v_br = 27mathrm~km/h$v_{br} = 27\mathrm{~km/h}$ at an angle of 150^circ$150^{\circ}$ with respect to the river flow (which is 30^circ$30^{\circ}$ upstream from the perpendicular crossing line). To find the width of the river, we only need the component of the boat's velocity perpendicular to the river bank (y$y$-axis): v_perp = v_br sin(150^circ) = 27 sin(150^circ) = 27 times frac12 = 13.5mathrm~km/h$$v_{\perp} = v_{br} \sin(150^{\circ}) = 27 \sin(150^{\circ}) = 27 \times \frac{1}{2} = 13.5\mathrm{~km/h}$$ Convert this velocity to SI units: v_perp = 13.5 times frac518mathrm~m/s = 3.75mathrm~m/s$$v_{\perp} = 13.5 \times \frac{5}{18}\mathrm{~m/s} = 3.75\mathrm{~m/s}$$ Given the crossing time is half a minute (t = 30mathrm~s$t = 30\mathrm{~s}$): w = v_perp cdot t = 3.75 times 30 = 112.5mathrm~m$$w = v_{\perp} \cdot t = 3.75 \times 30 = 112.5\mathrm{~m}$$ ### Step 1: Final Conclusion The width of the river is 112.5mathrm~m$112.5\mathrm{~m}$. ### Pattern Recognition In river-boat crossing problems, river velocity (v_r$v_r$) only causes drift along the bank; it has zero impact on crossing time or the crossing width calculations when the angle is defined relative to the flow. Use the vertical velocity component exclusively. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula v_textavg = fractextTotal DistancetextTotal Time$$v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}}$$ ### Core Logic Let's find the time taken for each section of the motion: - First section of distance x$x$ with velocity v_1 = 5mathrm~m/s$v_1 = 5\mathrm{~m/s}$: t_1 = fracxv_1 = fracx5$$t_1 = \frac{x}{v_1} = \frac{x}{5}$$ - Second section of distance frac32x$\frac{3}{2}x$ with velocity v_2$v_2$: t_2 = frac3x/2v_2 = frac3x2v_2$$t_2 = \frac{3x/2}{v_2} = \frac{3x}{2v_2}$$ Total distance is: d_texttotal = x + frac32x = frac52x$$d_{\text{total}} = x + \frac{3}{2}x = \frac{5}{2}x$$ Average velocity is: v_textavg = fracd_texttotalt_1 + t_2 = fracfrac52xfracx5 + frac3x2v_2$$v_{\text{avg}} = \frac{d_{\text{total}}}{t_1 + t_2} = \frac{\frac{5}{2}x}{\frac{x}{5} + \frac{3x}{2v_2}}$$ We are given v_textavg = frac507mathrm~m/s$v_{\text{avg}} = \frac{50}{7}\mathrm{~m/s}$. Equating the two values (noting x$x$ cancels out): frac507 = fracfrac52frac15 + frac32v_2$$\frac{50}{7} = \frac{\frac{5}{2}}{\frac{1}{5} + \frac{3}{2v_2}}$$ Divide both sides by 5: frac107 = fracfrac12frac15 + frac32v_2 implies 10 left(frac15 + frac32v_2right) = frac72$$\frac{10}{7} = \frac{\frac{1}{2}}{\frac{1}{5} + \frac{3}{2v_2}} \implies 10 \left(\frac{1}{5} + \frac{3}{2v_2}\right) = \frac{7}{2}$$ 2 + frac15v_2 = 3.5 implies frac15v_2 = 1.5 implies v_2 = frac151.5 = 10mathrm~m/s$$2 + \frac{15}{v_2} = 3.5 \implies \frac{15}{v_2} = 1.5 \implies v_2 = \frac{15}{1.5} = 10\mathrm{~m/s}$$ ### Step 1: Final Conclusion The velocity v_2$v_2$ is 10mathrm~m/s$10\mathrm{~m/s}$. ### Pattern Recognition Never take simple arithmetic averages of velocities! Average velocity must always be calculated as fractextTotal DistancetextTotal Time$\frac{\text{Total Distance}}{\text{Total Time}}$. Because total distance and time intervals are proportional to x$x$, x$x$ cleanly cancels out. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics

### Related Formula H = fracu^2 sin^2theta2g quad textand quad T = frac2u sinthetag$$H = \frac{u^2 \sin^2\theta}{2g} \quad \text{and} \quad T = \frac{2u \sin\theta}{g}$$ where, H$H$ = maximum height reached T$T$ = total time of flight u$u$ = initial projection velocity theta$\theta$ = angle of projection ### Core Logic From the formulas, we see that: H propto sin^2theta quad textand quad T propto sintheta$$H \propto \sin^2\theta \quad \text{and} \quad T \propto \sin\theta$$ Thus, we can directly link time of flight to the square root of the maximum height: T propto sqrtH implies fracT_1T_2 = sqrtfracH_1H_2$$T \propto \sqrt{H} \implies \frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}}$$ ### Step 1: Compute Ratio Given: H_1 = 8 H_2 implies fracH_1H_2 = 8$$H_1 = 8 H_2 \implies \frac{H_1}{H_2} = 8$$ Substitute this ratio: fracT_1T_2 = sqrt8 = 2sqrt2$$\frac{T_1}{T_2} = \sqrt{8} = 2\sqrt{2}$$ Thus, the ratio is 2sqrt2 : 1$2\sqrt{2} : 1$. ### Pattern Recognition Sees: Projectile heights ratio → Time of flight ratio. Shortcut: Since H propto u_y^2$H \propto u_y^2$ and T propto u_y$T \propto u_y$, we have T propto sqrtH$T \propto \sqrt{H}$. If the height is 8 times larger, the flying time is sqrt8 = 2sqrt2$\sqrt{8} = 2\sqrt{2}$ times larger. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinematics