A body of mass 250g is release d from the top of a building. If the body hits the ground with a velocity of 4m/s, calculate the height of the building (g = 10 m/s squared)​

the height of the building is 0.8 m.

Explanation:

Given;

mass of the object, m = 250 g = 0.25 kg

final velocity of the object, v = 4 m/s

let the height of the building = h

Apply the principle of conservation of energy to determine the height of the building;

¹/₂mv² = mgh

¹/₂v² = gh

Therefore, the height of the building is 0.8 m.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft