InterviewSolution
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InterviewSolution
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A gun is fired from a moving platform and ranges of the shot are observed to be `R_1 and R_2` when the platform is moving forwards and backwards, respectively, with velocity `v_P`. Find the elevation of the gun `prop` in terms of the given quantities. |
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Answer» Correct Answer - `tan^-1[(g)/(4 v_p ^2) xx (R_1 - R_2)^2/((R_1 + R_2))]`. Horizontal range `= (2 u_x u_y)/(g)` Let the initial horizontal and vertical components of the velocity of the shot be `u and v`, respectively, w.r.t platform. `R_1 = (2 (u + v_P)v)/(g)`, platform moving forward `R_2 = (2(u - v_P)v)/(g)`, platform moving backward `R_1 + R_2 = (4 u v)/(g) and R_1 - R_2 = (4 v_p v)/(g)` Now `(R_1 - R_2)^2 = (16 v_p ^2v^2)/(g^2)` and `((R_1 - R_2)^2)/(R_1 + R_2) = (4 x_p ^2)/(g) xx (v)/(u)` or `(v)/(u) = (g)/(4 v_p^2) ((R_1 - R_2)^2)/((R_1 + R_2))` Elevation of gun `prop = tan^-1((v)/(u)) = tan^-1 [(g)/(4 v_p^2) xx (R_1 - R_2)^2/((R_1 + R_2))]`. |
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