Question

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A stone of mass 10 Kg is dropped from a tower of height 122.5 m. After 1 second, if the gravitational force becomes zero. (Neglect air resistance) take g = 9.8ms^-1.

Find :

i) Velocity of the body after 2s is ______

ii) Neglecting the air resistance, the net force acting on the body is _____

iii) Distance covered in the 3rd second of its journey _____

Answer 1

Given,
A stone of mass 10 Kg is dropped from a tower of height 122.5 m
For first 1 second,
u=0

v=?

a=9.8m/sec²

t= 1

using the First equation of motion,
v= u + at

v= 0 + 9.8 *1

v= 9.8 m /sec

So Velocity after 1sec will be 9.8m/sec

((i) It is Given that after 1 second net gravitational force becomes zero,
So there will be no acceleration in body due to gravity,

Hence, body will fall with constant speed.
After 2 second velocity of body will remain 9.8m/sec

(ii) The force acting on body is given as product of mass and acceleration,
Net force' acting on body = m*a

= 10×9.8

= 98newton.
(iii)

Now ,
distance Covered in 3rd second:-
We have found that after 1 second body started moving with a constant velocity as Gravitational force becomes zero,
So body will cover 9.8 m in 3rd second,
Also using formula ,
distance Covered in nth second = u +a* 1/2(2n-)

putting a= 0, u= 9.8m/sec, n= 3

We get ,
Distance Covered in the
3rd second= 9.8m