# Difficult air resistance problem

1. Sep 25, 2007

### Xile

1. In a shotput event, a 2kg shotput is launched from a height of 1.5m with an initial velocity of 8m/s at an angle 60degrees to the horizontal. The formula R=0.0000378.v^2 gives the force due to air resistance (R) where v is the instantaneous velocity. Find the ratio of the gravitational force to the force due to air resistance, just before the shotput hits the ground.

If anyone could do this it would be much appreciated, cheers.

2. Sep 25, 2007

### Staff: Mentor

Please show some work. Write the equation of motions for horizontal and vertical flight with air resistance.

The force due to gravity is quite straightforward, but one needs the velocity to compute air resistance which will happen at an angle steeper than 60° to horizontal.

3. Sep 25, 2007

### Xile

I spent a while on it, and seemed to end up with some wierd looking equations. First i wrote an equation for both components of the velocity
Vv = Uv - (9.8 + Vv^2 . 0.0000378).t ie (0.0000378.t)Vv^2+Vv+(9.8t-Uv)=0
and same process (0.0000378t)Vh^2+Vh-Uh=0 keeping in mind we have the initial velocity and angle

i also got equations for displacement
Sv=Uvt - 0.5(9.8 + R).t^2
Sh=Uht - 0.5Rt^2
but i dont have the horizontal or vertical displacement, so i substituted into another equation and got
Vh^2 = Uh^2 - 2 . (9.8) . (Uht - 0.5Rt^2) which simplifies to
Vh=((16-8x9.8t)/(1-0.0000378x9.8)) but i dont have t, am i heading in the right direction coz i seem to be heading around in circles a bit.

4. Sep 25, 2007

### Xile

edit: that above equation is wrong and when i redid it its even more crazy with even mroe variables. i have the answer at the back of the book, its 8100:1 but i have no idea where it came from.

Last edited: Sep 25, 2007
5. Sep 25, 2007

### Staff: Mentor

Careful that R is a resistance force, so the acceleration would be R/m.

Also, with respect to the vertical velocity component, R acts downward with gravity as the shot ascends, but then acts opposite gravity when the shot descends.

R acts continuously against the horizontal velocity component.

6. Sep 26, 2007

### Xile

my mistake, but even so all i end up with is more variables and equations.

7. Sep 26, 2007

### Staff: Mentor

Just to be sure, R=0.0000378v2 is correct, right?

Then we know that R must always be less than 0.0000378 (8)2, because air resistance is always reducing the velocity components.

Treat vx and vy separately, and the vertical problem will have two parts.

Try using $$\frac{dv_x(t)}{dt}=-(k/m){v_x}^2(t)$$ for the horizontal motion.

vertically $$\frac{dv_y(t)}{dt}=-(k/m){v_y}^2(t)-g$$ on the way up. On the way down, change the sign in front of the air resistance term.

8. Sep 26, 2007

### Xile

ah, that looks a lot better but i dont really understand how you got to there? it confuses me..

9. Sep 26, 2007

### Staff: Mentor

I simply wrote the equations for the changes in velocity (acclerations) in the vertical and horizontal. I applied Newton's laws, and then divide the forces by m to get accelerations.

The change in horizontal velocity is strictly due to air resistance, which is strictly related to the horizontal velocity.

I did it similarly for the vertical direction, but there gravity is always acting downward, and realize that the air resistance operates against the velocity - down going up and up going down. At the apogee, the problem becomes one of free fall, but with air resistance.

10. Sep 26, 2007

### Xile

but wouldnt the Vx(t)^2 be in the numerator, same with Vy(t)^2

11. Sep 26, 2007

### Staff: Mentor

No, when rearranging, one obtains dv/v2 = -k/m dt in x, and

dv/(g+(k/m)v2) = - dt in y for the way up.