A metal box attached to a small parachute is dropped from helicopter

A.explain in terms of the forces acting on why its velocity increased immediately after being dropped

B.why did it reach a uniform velocity after a short distance

C.the total force opposing motion of the box and the parachute at particular time during its fall is 30N.the combined mass of the box and parachute is 5kg
1. Caculate the resultant downward force and box and the parachute (g=10N)
2.briefly describe the motion of the box and parachute at this time

Can anyone plz tell me from which book did u got this question

Immediately after drop its vertical speed is zero, so no drag.

therefore m a = m g and it accelerates down at 32 m/s^2
as it builds up speed and the chute deploys, the air drag increases at about proportional to speed squared until the drag force up equals the m g down and the speed becomes constant.
I already did your 30 N and g = 10 METERS/ SECOND^2 (not Newtons) question

A metal box attched to a small parachute is dropped from helicopter

a) explain in term of force acting why

A. When the metal box attached to a small parachute is dropped from a helicopter, initially, there are two main forces acting on it: gravity and air resistance. Gravity pulls the box downward with a force equal to its weight, which is the mass of the box multiplied by the acceleration due to gravity. Air resistance, on the other hand, acts in the opposite direction to the motion of the box, and it increases with the speed of the falling object.

Since the box is initially at rest when dropped, the force of gravity is greater than the air resistance, causing the box to start accelerating downward. As the box gains velocity, the air resistance becomes stronger, but it is still lower than the force of gravity at this point. Therefore, the net force acting on the box is downward, resulting in an increase in velocity immediately after being dropped.

B. After a short distance, the box and parachute will eventually reach a uniform or terminal velocity. This happens when the air resistance becomes equal to the force of gravity acting on the box. At terminal velocity, the net force becomes zero, and there is no acceleration anymore. The box continues to fall, but its velocity remains constant.

Initially, the box accelerates due to the force of gravity, but as it gains speed, the air resistance increases. Eventually, the air resistance becomes strong enough to balance the force of gravity, resulting in a uniform velocity.

C.
1. To calculate the resultant downward force on the box and parachute, we need to subtract the opposing force (30N) from the gravitational force. The gravitational force can be calculated using the formula:
Gravitational force = mass * acceleration due to gravity

Gravitational force = 5kg * 10N/kg = 50N

Resultant downward force = Gravitational force - Opposing force
= 50N - 30N
= 20N

2. At this time, when the resultant downward force is 20N, the motion of the box and parachute can be described as falling with a constant velocity. The opposing force (air resistance) is now equal to 30N, and the gravitational force acting on the box is 50N. Since the opposing force is not strong enough to oppose the gravitational force fully, there is still a net downward force of 20N, causing the object to continue falling. However, since the net force is constant, the object falls at a constant velocity, without accelerating further.