Kinematics Problem on projectile

1. May 15, 2008

jason_mcc

1. The problem statement, all variables and given/known data

A 10 kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s. Determine the maximum height to which it will travel if (a) atmospheric resistance is neglected, and (b) if atmospheric resistance is measured as Fd=(0.01 v2), where v is the speed at any instant, measured in m/s.

2. Relevant equations

v$$^{2}$$=u$$^{2}$$+2as

3. The attempt at a solution

Part (a) I know how to do.

v$$^{2}$$=u$$^{2}$$+2as
0=50$$^{2}$$+2(-9.8)s
0=2500-19.6s
s=127.55m

But I can't do part (b). I don't know how to apply the air resistance to the answer.

Can anyone help?

2. May 15, 2008

Hootenanny

Staff Emeritus
Welcome to PF jason,

For part (b) the resistive force and hence acceleration is not constant. Therefore, you cannot use equations of constant acceleration. Instead, you should consider Newton's Second Law.

3. May 15, 2008

jason_mcc

For the equation

F=0.01v$$^{2}$$

which value for v would I use to find the air resistance?

4. May 15, 2008

Hootenanny

Staff Emeritus
Dont get ahead of yourself. Start by writing out an equation of motion using Newton's second law.

5. May 15, 2008

jason_mcc

Ok so far iv got

F=98.1+0.01v$$^{2}$$
ma=98.1+0.01v$$^{2}$$
10a=98.1+0.01v$$^{2}$$ since mass is constant
a=9.81+0.001v$$^{2}$$
$$\stackrel{dv}{dt}$$=9.81+0.001v$$^{2}$$

Am I on the right lines?

6. May 15, 2008

Hootenanny

Staff Emeritus
You're on the right lines but you need to be careful the way you define your coordinate axis and velocity. Notice that both the weight of the projectile (mg) and the drag are acting against the motion of the projectile. Hence, your ODE should really be

$$\frac{dv}{dt} = - \left(g+\frac{v^2}{1000}\right)$$

So, you now have a first order ODE. Can you solve it?

7. May 16, 2008

jason_mcc

Is it

v=-9.81t-$$\frac{v^{3}}{3000}$$$$\frac{dv}{dt}$$

And then i sub back in for dv/dt?

8. May 16, 2008

DavidWhitbeck

Jason you are mixing up your calculus rules. You can't evaluate $$\int v^2 dt$$ without knowing the solution-- how v depends on time.

You must separate and integrate. Get all your v's on the left hand side and all of your t's on the right side and then integrate.

9. May 16, 2008

jason_mcc

Ok is it

t=ln|-9.81-$$\frac{v^{2}}{1000}$$| ?

10. May 16, 2008

Hootenanny

Staff Emeritus
No, let's do this one step at a time. Start by separating it and show me what you get.

11. May 16, 2008

jason_mcc

Ok

$$\frac{dv}{dt}$$=-(9.81+$$\frac{v^{2}}{1000}$$)

dv=(-9.81-$$\frac{v^{2}}{1000}$$)dt

$$\frac{dv}{-9.81-v^{2}/1000}$$=dt

12. May 16, 2008

DavidWhitbeck

Jason, I suggest using partial fractions on that integral.

13. Jun 8, 2008