Tossing upward with air resistance

[jvileisis]

Could someone tell me if my approch to this is correct? I am given the change in vertical v final(zero), v initial, mass, and frictional coefficient. The question askes for distance. I set up the equation vf^2=vi^2+2ad(mg(frictional coefficent))

 

[G01]

I don't understand what you mean by that equation. If you are multiplying the 2ad term by the force of friction from the air then you are incorrect.

That kinematic equation is still valid in its normal form. The difference is that "a" will not be the acceleration caused by gravity, but it will be the acceleration caused by the sum of gravity and the drag force.

HINT: Can you find this acceleration starting from Newton's Second Law?

 

[ogb p]

Your approach to this problem appears to be correct. To find the distance, you can use the equation d = (vf^2 - vi^2) / (2a), where d is the distance, vf is the final velocity, vi is the initial velocity, and a is the acceleration. In this case, the acceleration would be equal to g (gravity) minus the frictional force (mg(frictional coefficient)). This will give you the distance traveled while taking into account the air resistance.