Q27 of 35 Page 55

A body is projected from height 60m with a velocity 10 ms–1 at angle 30° to horizontal. The time of flight of the body is? (g = 10ms–2)


Here u = 10 ms–1 and Ɵ = 30°


Now, let us consider the vertical parameters:



The net vertical distance covered by the particle


= h – h – 60


Or, s = –60 m


The negative sign is considered as the measurement was taken in the negative y-direction.


Now, acceleration = –10 ms–2


So, using the equations of motion, we get:






So, t = 4 s and thus the time of flight is 4 s.


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