Question

A particle is thrown vertically up with initial
velocity of 60 m/s. The distance covered by the
particle in first two seconds of descent will be
(take g = 10 m/s2)
(1) 5 m
(2) 15 m
(3) 20 m
(4) 40 m
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Answer 1

hence distance is 20m. hope it will work thanks. please mark me brainlist.

Answer 2

Answer:20m

Explanation: distance covered( from the frame of reference of the particle) during descent doesn't depend on the initial velocity of the particle in the positive y axis direction; it depends upon gravity. Taking g as 10m/s^2, we use the formula S=0.5g×t^2 to get S=20m