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Kinematics Problem on projectile

  • Thread starter jason_mcc
  • Start date
  • #1
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Hi everyone, any help you could give me I would appreciate.

Homework Statement



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.


Homework Equations



v[tex]^{2}[/tex]=u[tex]^{2}[/tex]+2as

The Attempt at a Solution



Part (a) I know how to do.

v[tex]^{2}[/tex]=u[tex]^{2}[/tex]+2as
0=50[tex]^{2}[/tex]+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?
 

Answers and Replies

  • #2
Hootenanny
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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
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For the equation

F=0.01v[tex]^{2}[/tex]

which value for v would I use to find the air resistance?
 
  • #4
Hootenanny
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For the equation

F=0.01v[tex]^{2}[/tex]

which value for v would I use to find the air resistance?
Dont get ahead of yourself. Start by writing out an equation of motion using Newton's second law.
 
  • #5
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Ok so far iv got

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

Am I on the right lines?
 
  • #6
Hootenanny
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Ok so far iv got

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

Am I on the right lines?
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

[tex]\frac{dv}{dt} = - \left(g+\frac{v^2}{1000}\right)[/tex]

So, you now have a first order ODE. Can you solve it?
 
  • #7
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Is it

v=-9.81t-[tex]\frac{v^{3}}{3000}[/tex][tex]\frac{dv}{dt}[/tex]

And then i sub back in for dv/dt?
 
  • #8
Jason you are mixing up your calculus rules. You can't evaluate [tex]\int v^2 dt[/tex] 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
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Ok is it

t=ln|-9.81-[tex]\frac{v^{2}}{1000}[/tex]| ?
 
  • #10
Hootenanny
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Ok is it

t=ln|-9.81-[tex]\frac{v^{2}}{1000}[/tex]| ?
No, let's do this one step at a time. Start by separating it and show me what you get.
 
  • #11
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Ok

[tex]\frac{dv}{dt}[/tex]=-(9.81+[tex]\frac{v^{2}}{1000}[/tex])

dv=(-9.81-[tex]\frac{v^{2}}{1000}[/tex])dt

[tex]\frac{dv}{-9.81-v^{2}/1000}[/tex]=dt
 
  • #12
Jason, I suggest using partial fractions on that integral.
 
  • #13
hey can anyone help me with this?

A hot air balloon is descending at a rate of 2.0 m/s. when a passenger drops a camera If the camera is 45 m above the ground when is it dropped? how long does it take for the camera to reach the ground< and (b) what is its velocity just before it lands? Let upward be the positive direction for this problem.

i know this is the formula to start out with ...
X=Xo+Vot+ 1/2at^2
45=o+(.5)(9.81m/s.)t
 
Last edited:

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