An engine approaches a high vertical wall with constant speed of 90 m/sec. When it is at a distance of 1050 m from the wall, it blows a whistle, whose echo is heard after 5 second. Calculate the speed of sound in air.

Velocity with which the engine is approaching the wall = 90 m/s
Let V be the velocity of sound and u be the velocity of the source.
Echo is heard after time t.
Let the engine whistle when it is at P and hear the echo at Q. 
 
∴   Distance travelled by engine in time t. 
                 PQ = ut                      ...(1) 
Distance travelled by sound in time t. 
                 PR + RQ = Vt             ...(2) 
On adding (1) and (2), we get
            PR + RQ + PQ = 2PR
                                  = (V+u)t 
Here, we have
Initial velocity, u = 90 m/s
Distance travelled, PR = 1050 m
Time taken to hear the echo, t = 5s 
Therefore, 
Speed of sound in air is given by, 
           2 x 1050 = (V + 90)x 5
                 5V = 1650
                   V = 330 m/s