Sky Diver Physics: Acceleration and Air Resistance Calculation

[lmf22]

(a) What is the acceleration of two falling sky divers (mass 110.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight?

(b) After popping open the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute?

any help would be great. I don't know where to start.

 

[Leong]

Upward direction positive; downward direction negative
$$\begin{array}{cc} Newton's \ 2nd\ Law\\ \sum \vec{F}=m\vec{a}\\ W+D=ma\\ -mg + 0.25mg = ma\\ \end{array}$$

Part (b)
1.Draw the free body diagram for the divers and assign all the forces acted on them.
2. Use Newton's 2nd Law.
3.What is the acceleration when they are moving with constant velocity ?
4.Solve the equation.
4.Becareful with the signs.

 

[M98Ranger]

(a) To calculate the acceleration of the sky divers, we can use the equation F = ma, where F is the net force acting on the divers, m is their mass, and a is their acceleration. We know that the upward force of air resistance is equal to one-fourth of their weight, so we can set up the following equation:

F = 1/4mg = ma

Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2. We can rearrange this equation to solve for a:

a = F/m = (1/4mg)/m = 1/4g

Plugging in the mass of the divers (110.0 kg), we get an acceleration of approximately 2.45 m/s^2.

(b) After the parachute is opened, the divers are now descending at a constant speed, which means their acceleration is zero. This also means that the net force acting on them is also zero. The force of air resistance on the divers and their parachute is equal and opposite to the force of gravity pulling them down. So, the force of air resistance is now equal to their weight, which is approximately 1100 N (110.0 kg x 9.8 m/s^2). This is because the parachute increases the surface area of the divers, increasing the force of air resistance acting on them.