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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 _____
PLEASE GIVE CORRECT ANSWER AND GIVE FULL STEPS. PLZZZZZ I HAVE EXAM TOMORROW. HELP ME. ANSWER WILL BE BRAINLIEST
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 _____
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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