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A jet airplance travelling at the speed of ` 500 km ^(-1)` ejects its products of combustion at the speed of ` 1500 km h^(-1)` relative to the jet plane. What is the speed of the later with respect to observer on the ground ? |
Answer» Let (v -p)` be the velocity of the products w.r.t. ground, Let us consiedr the direction of moton of airplane to be positive direction of X-axis. Here, speed of jet plane, ` v_A =500 km h^(-1)` Relative speed of products of combustion w.r.t to jet plane , v_(pA0 = - 1500 =- 1000 km h^(-10` Relative velocity of the projucts w.r.t jet plane is ` v_(PA) = v_(p) =- v_A =- 1500 or v_p =v _A- 1500 =500 - 1500 =- 1000 km h^()-1)` Here- ve sign shows that the direction of products of combustion os opposite to tht of the airplane. Thus the magnitude of relative velocity is ` 1000 km h^(-1)` . |
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