Question

A particle is thrown with any velocity vertically upward,the distance travelled by the particle in 1st second of it's decent is-
a.g/2
b.g
c.g/4
d.can't be calculated

Answer 1

JUST CONSIDER THIS CONCEPT THAT BEFORE IT'S MOTION OF DESCENT IT WILL REACH THE HIGHEST POINT .

AND AT HIGHEST POINT IT'S VELOCITY WILL
BE ZERO. THIS WILL BE THE SAME CASE, AS IF THE PARTICLE IS DROPPED FROM A HEIGHT WITH IT'S INITIAL VELOCITY ZERO.

NOW USE,,
s=ut+1/2at^2
AS U=0 ,
SO, s==1/2gt^2
s=1/2 × g
s=g/2.


HOPE U FIND THIS HELPFUL. ☺️☺️☺️☺️

Answer 2

Answer:

Explanation:

USE,,

s=ut+1/2at^2

AS U=0 ,

SO, s==1/2gt^2

s=1/2 × g

s=g/2.

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