Question:
a jet aeroplane travelling at the speed of 500 km per hour ejects its products of combustion at the speed of 1500 km per hour relative to the jet plane . what is the speed of the latter with respect to the observer on the ground
Answer:
Let the velocity of the jet = Vj
Let the velocity of the ejected gas = Vg
And let the velocity of the observer on the ground = V0
Now, speed of the jet relative to the observer on ground = 500 km/hr
Therefore, Vj - V0 = 500 km/hr (Equation 1)
Assume, V0 = 0
Now since the speed of the ejected gases relative to jet = 1500 km/hr, which is in the opposite direction to the movement of the jet,
So, Vg - Vj = - 1500 km/hr (negative since it is in opposite direction) (Equation 2)
Adding up equations 1 and 2,
We get, Vg -V0 = 1000 km/hr.
Thus, speed of the ejected products with respect to the observer on ground is 1000 km/hr.
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