A jet airplane travelling at the speed of 500 kmh-1 ejects the burnt gases at the speed of 1400 kmh-1 relative to the jet airplane. The speed of burnt gases relative to stationary observer on the earth is

(A) 2.8 kmh-1

(B) 190 kmh-1

(C) 700 kmh-1

(D) 900 kmh-1


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Correc Option is (D) 900 \(kmh^{-1}\)

\(v_p = -500\) km/h

\(V_{gp} = 1400\) km/h

The speed of burnt gas with respect to the jet airplane is

\(\overrightarrow v_{gp} = \overrightarrow v_g - \overrightarrow v_p\)

\(1400 = v_g + 500\)

\(v_g = 1400 - 500\)

\(v_g = 900\) km/h

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Correct option is: (D) 900 kmh-1

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