Match column type question At any instant, the horizontal position to balloon is defined by \(\mathrm{x}=\mathrm{gt} \mathrm{m} .\) Equation of path is \(\left(\mathrm{x}^{2} / 30\right)\). After \(\mathrm{t}=2 \mathrm{sec}\), match the following. Table 1 Table 2 (A) distance of balloon from station \(\mathrm{A}\). (P) \(14.1,50^{\circ}\) (B) magnitude \& direction of velocity (Q) \(24 \mathrm{~m}\) C) magnitude \& direction of acceleration (R) \(21 \mathrm{~m}\) (S) \(5.4,90^{\circ}\) (A) \(\mathrm{A}-\mathrm{R}, \mathrm{B}-\mathrm{S}, \mathrm{C}-\mathrm{P}\) (B) \(\mathrm{A}-\mathrm{Q}, \mathrm{B}-\mathrm{S}, \mathrm{C}-\mathrm{P}\) (C) \(\mathrm{A}-\mathrm{R}, \mathrm{B}-\mathrm{P}, \mathrm{C}-\mathrm{S}\) (D) \(\mathrm{A}-\mathrm{Q}, \mathrm{B}-\mathrm{P}, \mathrm{C}-\mathrm{S}\)

Step by step solution

01

Find the horizontal position x

At any instant, the horizontal position x of the balloon is described by the equation \(x = gtm\). Given that \(t = 2s\), with \(g = 10 m/s^2\), we have: \[x = 10 \times 2m = 20m\]

02

Find the vertical position y

To obtain the vertical position of the balloon at \(t = 2s\), we use the given equation for the path of the balloon: \[y = \frac{x^2}{30}\] Substitute the value of x found in Step 1: \[y = \frac{(20)^2}{30} = \frac{400}{30} = 13.33m\]

03

Determine the distance of the balloon from the station

Now, we compute the distance (A) of the balloon from the station A using the Pythagorean theorem: \(A = \sqrt{x^2 + y^2}\) Substitute the values of x and y found in Steps 1 and 2: \[A = \sqrt{(20)^2 + (13.33)^2} = \sqrt{400 + 177.77} ≈ 24m\] Verify the correct match from Table 1 and 2: A matches with Q.

04

Compute the magnitude and direction of the velocity (v_x and v_y)

Given \(x = gtm\), we have: \[\frac{dx}{dt} = v_x = gt\] The horizontal velocity \(v_x = 10 \times 2 = 20m/s\) Since the balloon moves vertically following a parabolic path, we can differentiate the equation \(y = \frac{x^2}{30}\) with respect to time t: \[\frac{dy}{dt} = \frac{d}{dt} [\frac{x^2}{30}]\] Since \(\frac{dx}{dt} = v_x\), we get: \[\frac{dy}{dt} = \frac{2x \times v_x}{30}\] Substituting the values of x and \(v_x\), we get: \[\frac{dy}{dt} = v_y = \frac{2 \times 20 \times 20}{30} = 26.67 m/s\] Now, compute the magnitude of the velocity (B): \[v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20)^2 + (26.67)^2} ≈ 33.33 m/s\] Next, find the direction of the velocity in degrees: \[\theta = \tan^{-1}(\frac{v_y}{v_x}) = \tan^{-1}(\frac{26.67}{20}) ≈ 53.33^{\circ}\] Verify the correct match from Table 1 and 2: B matches with P.

05

Calculate the magnitude and direction of the acceleration (a_x and a_y)

In the horizontal direction, \(a_x = 0\) (constant horizontal velocity) In the vertical direction, we second differentiate the equation \(y = \frac{x^2}{30}\): \[\frac{d^2y}{dt^2} = \frac{d}{dt} [\frac{2x \times v_x}{30}]\] Since \(a_x = 0\), we have: \[\frac{d^2y}{dt^2} = a_y = \frac{2 \times v_x^2}{30}\] Substituting the value of \(v_x\): \[a_y = \frac{2 \times 20^2}{30} = 26.67 m/s^2\] Compute the magnitude of the acceleration (C): \[a = \sqrt{a_x^2 + a_y^2} = \sqrt{(0)^2 + (26.67)^2} = 26.67m/s^2\] Then, find the direction of the acceleration in degrees: \[\alpha = \tan^{-1}(\frac{a_y}{a_x}) \] Since \(a_x = 0\), we have: \[\alpha = 90^{\circ}\] Verify the correct match from Table 1 and 2: C matches with S. Based on our calculations and matches: The correct answer is (D): A-Q, B-P, C-S.

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