Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity

on an object falling we have three forces acting mainly

force of gravity : acts in downward direction

force of buoyancy : acts in upward direction

drag force due to air : acts in upward direction

the force of gravity makes the object fall in downward direction and force of buoyancy acts in opposite upward direction to oppose the motion. initially since force of gravity was greater than the force of buoyancy , the net force is acting in down direction and hence the object gains speed in downward direction.

on the object falling down , drag force by air acts in upward direction to oppose is downward motion. the drag force is directly propotional to the square of the speed. hence as the object gains speed in down direction , the drag force increase in upward direction.

at some velocity called terminal velocity , the sum of drag force and force of buoyancy becomes equal to the force of gravity  and then there is no net force on the object . so the object then moves at constant terminal velocity.

at terminal velocity ,

force of gravity = force of air resistance + force of buoyancy

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