Question

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 isii) Neglecting the air resistance, the net force acting on the body isiii) Distance covered in the 3rd second of its journey

Answer 1

i) Velocity of the body after 2s is -9.8m/s.

ii) Neglecting the air resistance, the net force acting on the body is 98N in the first second of its journey and 0N for the rest of its journey.

iii) Distance covered in the 3rd second of its journey is 9.8 m.

Given:

Mass of stone = 10kg

Height of tower = 122.5m

Gravitational force after 1s = 0N

Gravitational acceleration = 9.8m/s

To find:

i) Velocity of the body after 2s

ii) The net force acting on the body

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

Answer:

i) Velocity of the body after 2s

• For the first second of the stone's journey-
• We know that, v= u+at
• v= 0 + (-9.8)*1
• v = -9.8 m/s
• This velocity becomes the initial velocity for the stone during its 2nd second of journey.
• From, v= u+at
• v = -9.8 + (0)*1
• v = -9.8 m/s
• The negative sign indicates that the velocity is directed towards the ground.

ii) The net force acting on the body

• For the first second of the jouney, from the equation
• F = ma
• We have, F = 10 * (-9.8) = -98N
• From the 2nd second onwards, since the gravitational force becomes zero the net force acting on the stone is 0N.

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

• For the third second of the journey, the initial velocity would be -9.8m/s.
• From, s = ut + 1/2 at²
• s = -9.8(1) + 1/2 (0)(1)²
• s = -9.8 m

• Here, the negative sign indicates that the displacement is directed towards the ground.

• Thus the distance covered = 9.8m