Question

A particle is thrown verticlally up with initial velocity 60m/s the distance cover by the particle in first 2 seconds of descent will be​

Answer 1

Explanation:

v = 0

u = 60 m/s

t = 2

v = u + at

0 =60 +2a

-30 = a

v2 - u2 = 2as

0 - 60 × 60 = 2(-30)s

-60 × 60 = -60s

s = 60 m

Answer 2

Answer:

v = 0

u = 60 m/s  

t = 2

v = u + at

0 =60 +2a  

-30 = a  

v2 - u2 = 2as  

0 - 60 × 60 = 2(-30)s  

-60 × 60 = -60s  

s = 60 m

Explanation:

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