Question # A jet airplane travelling from east to west at a speed of ( 500 mathrm{km} mathrm{h}^{-1} ) ejected out gases of combustion at a speed of ( 1500 mathrm{km} mathrm{h}^{-1} ) with respect to the jet plane. What is the velocity of the gases with respect to an observer on the ground?

# A jet airplane travelling from east to west at a speed of ( 500 mathrm{km} mathrm{h}^{-1} ) ejected out gases of combustion at a speed of ( 1500 mathrm{km} mathrm{h}^{-1} ) with respect to the jet plane. What is the velocity of the gases with respect to an observer on the ground?

(A) ( 1000 mathrm{km} mathrm{h}^{-1} ) in the direction west to east

(B) ( 1000 mathrm{km} mathrm{h}^{-1} ) in the direction east to west

(C) ( 2000 mathrm{km} mathrm{h}^{-1} ) in the direction west to east

(D) ( 2000 mathrm{km} mathrm{h}^{-1} ) in the direction east to west

Solution

Velocity of jet w.r.t. ground ( mathbf{V}_{mathrm{jg}}=500 mathrm{km} / mathrm{h} )

(Upward) Velocity of products relative to jet ( mathbf{V}_{mathrm{P} mathrm{J}}=1500 ) ( mathrm{km} / mathrm{h} )

Hence, velocity of products relative to ground = ( -1500+500 )

[

=

]

( -1000 mathrm{km} / mathrm{h} )

Here - ve sing means downward.