Stone \(\mathrm{A}\) is thrown in horizontal direction with velocity of $10 \mathrm{~ms}^{-1}\( at the same time stone \)\mathrm{B}$ freely falls vertically in downward direction. Calculate the velocity of \(\mathrm{B}\) with respect to \(\mathrm{A}\) after 10 second. (A) \(10 \mathrm{~ms}^{-1}\) (B) \(\sqrt{(101)} \mathrm{ms}^{-1}\) (C) \(10 \sqrt{(101)} \mathrm{ms}^{-1}\) (D) 0

Given data: - Stone A is thrown horizontally with a velocity of 10 m/s - Stone B falls freely vertically downward Relevant formulas: 1. Final velocity after time t: \(v_f = v_i + at\) 2. Relative velocity: \(v_{rel} = v_B - v_A\)

The horizontal velocity of stone A remains constant because there is no acceleration in the horizontal direction. Therefore, the horizontal velocity of stone A after 10 seconds is 10 m/s.

Since stone B is falling freely, it experiences acceleration due to gravity, which we will take as \(9.8\mathrm{~ms}^{-2}\). Using the given time (10 seconds), we can calculate the final vertical velocity of stone B after 10 seconds using the formula: \(v_f = v_i + at\) Since stone B is falling freely, we take \(v_i = 0\mathrm{~ms}^{-1}\). \(v_f = 0 + (9.8)(10)\) \(v_f = 98\mathrm{~ms}^{-1}\) Stone B has a vertical velocity of 98 m/s after 10 seconds.

To find the relative velocity of stone B with respect to stone A, we will have to take vector differences between the horizontal and vertical directions. Let vx represent the horizontal component and vy represent the vertical component of the velocities: \(v_{Bx} = 0 \mathrm{ms}^{-1}\) (Because Stone B falls vertically) \(v_{By} = 98 \mathrm{~ms}^{-1}\) \(v_{Ax} = 10 \mathrm{~ms}^{-1}\) \(v_{Ay} = 0 \mathrm{~ms}^{-1}\) (Because Stone A moves horizontally) Now we calculate the relative velocity components: \(\Delta v_x = v_{Bx} - v_{Ax} = 0 - 10 = -10\mathrm{~ms}^{-1}\) \(\Delta v_y = v_{By} - v_{Ay} = 98 - 0 = 98\mathrm{~ms}^{-1}\) Now we calculate the magnitude of the relative velocity: \(v_{rel} = \sqrt{(\Delta v_x)^2 + (\Delta v_y)^2}\) \(v_{rel} = \sqrt{(-10)^2 + (98)^2}\) \(v_{rel} = \sqrt{100 + 9604}\) \(v_{rel} = \sqrt{9704}\) \(v_{rel} = 10\sqrt{101} \mathrm{~ms}^{-1}\) The velocity of stone B with respect to stone A after 10 seconds is \(10\sqrt{101} \mathrm{~ms}^{-1}\). The correct answer is (C) \(10\sqrt{101} \mathrm{~ms}^{-1}\).