Question

Heya❤️❤️❤️

good evening...

please answer..


dont spamm❎❎❎❎


(^_-)-☆


a balloon starts rising from ground from rest with an upward acceleration of 2m/^2.just after 1 second,a stone is dropped from it .the time taken by stone to strike the ground is nearly-

(^_-)-☆

fast please....

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Answer 1

Explanation:

GIVEN,

A=2M/S ^2

SEE ITS GIVEN THAT IT STARTS FROM THE REST THEREFORE ITS INITIAL POINT (u)=0

T=1S

LET'S FIND ITS INITIAL VELOCITY NOW,

V=U+AT

V=0+2×1=2M/S

NOW,

THE DISTANCE,

V^2=U^2as

V^2=2as

•°• S=V^2/2a

=2^2/2×2

=4/4=1

HENCE,THE STONE FALLS 1M BEFORE HITTING THE GROUND.

NOW,

THE ACCELERATION AFTER BEING DROPPED,

U=-2m/s

a=9.8m/s^2

t=1s

NOW,BY THIRD EQUATION OF MOTION WE CAN FIND THE TIME WHICH IS OUR ANSWER.....

s=ut+1/2 at^2

1=-2t+1/2×9.8^2

4.9t^2-2t-1=0

t=2+- root (-2)^2-4×4.9×(-1)/2×9.8

=2root 4+19.6/9.8

=2root23.6/9.8

t=0.7s [finally I got it]

HENCE,THE STONE WILL STRIKE THE GROUND IN 0.7s.