Question

3.
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

answer : option (3) 20m

this question is little different. question want to find distance covered by particle in first two seconds of descent.

particle first moves upward, after reaching a specific height, it falls downward (descending motion). here observing point is that at specific height velocity of particle becomes zero after then it will ready to fall.

so, initial velocity of particle in descending motion, u = 0

distance travelled by particle , s = ut + 1/2 at²

here, u = 0, t = 2s and a = -g = -10m/s²

then, s = 0 + 1/2 (-10m/s²) × (2s)²

= -20m [ here negative sign indicates that particle is moving downward ]

hence, distance covered by particle in first two seconds of descent will be 20m

Answer 2

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