A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms\(^{-2}\)]. Calculate the speed of projection
1.2ms-1
6.0ms-1
12.0ms-1
0.6ms-1
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The answer is 6m/s. Its time of flight was 1.2s, that means the time it took for it to reach max height is half that time. And only at max height is final velocity v = 0m/s. We can use v = u - gt where v = 0, g = 10m/s2 and t = 0.6s.
Time of flight (T) is the time required for a projectile to return to the same level from which it was projected.
T = (2u*sinϴ)/g
since it is projected vertically upwards ϴ = 90
T = 1.2
1.2 = (2u*sin90)/10
12 = 2u
u = 6
The time taken for the body of mass to move from the point of projection and come back is the same as the time taken for the body of mass to move from the point of projection to the point of landing (Time Of Flight).
But the time taken to reach maximum height is time of flight divided by 2 which is 1.2/2 = 0.6
