Solving ODE to find general solution

In summary: If he initially starts at 4000m, then wouldn't y(0)=4000? What is y(0) equal to from your initial conditions when solving the ODE? Similarly the diver has an initial velocity so y'(0)=0. What y'(0) equal to?And when the diver hits the ground, his displacement 'y should be zero.Ok so if that is the case y(0)= 4000 would give me an answer of, 4000 = y0 which was the initial condition. The second was that y' = v0. So Substituting back into general solution would give, y = 4000 + (0)t- 1/2gt^2, where initial vertical
  • #1
476
0
If we assume air resistance is negligible, the only force acting on a body is -Mg where g is the acceleration due to gravity ( negative because acting downwards). F = Ma becomes :
-Mg = M y''
which implies
y''=-g

Question asks find the general solution for y.



Homework Equations


y is the distance above the ground of an object
y' is the vertical velocity
y'' is vertical accelaration


The Attempt at a Solution



Here is what I have done so far:

y'' = -g, therefore y' = -gt + A
integrating again
y= -1/2 gt^2 + At + B

Am I correct so far or do I take gravity to be a constant such as x giving me an answer of:

y = -3/2 g^3 + At + B
 
Physics news on Phys.org
  • #2
Jamiey1988 said:

The Attempt at a Solution



Here is what I have done so far:

y'' = -g, therefore y' = -gt + A
integrating again
y= -1/2 gt^2 + At + B

Am I correct so far or do I take gravity to be a constant such as x giving me an answer of:

y = -3/2 g^3 + At + B

No no, the first one is quite correct. Acceleration due to gravity is constant. So your first general solution is correct.
 
  • #3
Thanks for that, following on the question asks to find particular solutions for:
y(0) = [y][0] and y'=[v][0] for which i get:

[y][0] = B and [v][0] = A - gt

Are these correct also?
 
  • #4
Jamiey1988 said:
Thanks for that, following on the question asks to find particular solutions for:
y(0) = [y][0] and y'=[v][0] for which i get:

[y][0] = B and [v][0] = A - gt

Are these correct also?

Yes those are correct. I assumed by [y][0] you meant y0 such that y(0)=y0 and not y(0)= 0.
 
  • #5
Ye sorry that is what I meant. Now is the part I am really stuck with. The final part pf the question states,
A sky diver whose mass is 80kg leaps from a plane at 4000m above the ground and his parachute fails to open. If initial velocity is zero at what time does he hit the ground? and how fast is he going when he hits? Assume g = 10m.s^-2.
I am not sure as to tackle this question, maybe using basic formula such as v^2 = u^2 + 2as something like that?
 
  • #6
Jamiey1988 said:
Ye sorry that is what I meant. Now is the part I am really stuck with. The final part pf the question states,
A sky diver whose mass is 80kg leaps from a plane at 4000m above the ground and his parachute fails to open. If initial velocity is zero at what time does he hit the ground? and how fast is he going when he hits? Assume g = 10m.s^-2.
I am not sure as to tackle this question, maybe using basic formula such as v^2 = u^2 + 2as something like that?

Ok, well you just solved the equation y''=-g and got y=y0+v0t -1/2gt2 right?

If he initially starts at 4000m, then wouldn't y(0)=4000? What is y(0) equal to from your initial conditions when solving the ODE? Similarly the diver has an initial velocity so y'(0)=0. What y'(0) equal to?

And when the diver hits the ground, his displacement 'y should be zero.
 
  • #7
Ok so if that is the case y(0)= 4000 would give me an answer of, 4000 = y0 which was the initial condition. The second was that y' = v0. So Substituting back into general solution would give,
y = 4000 + (0)t- 1/2gt^2, where initial vertical velocity is zero, v0 = 0
So from there 4000 = 1/2 gt^2
8000 = gt^2
t^2 = 8000/10 (As gravity = 10m.s^-2)
t = SQRT (800)
t = 28.284 seconds
Correct?
 
  • #8
Yes that looks correct.
 
  • #9
Thanks for all your help :smile:
 
  • #10
Oops sorry I forgot to answer the speed at which he hits the ground? U say his displacement 'y should be zero. could u expand on that for me please.
 

1. What is an ODE?

An ODE, or ordinary differential equation, is a mathematical equation that describes how a function changes over time. It involves an unknown function and its derivatives.

2. How do you solve an ODE?

To solve an ODE, you need to find the general solution, which is the set of all possible solutions to the equation. This can be done using various methods such as separation of variables, integrating factors, or using series solutions.

3. What is a general solution?

A general solution is the set of all possible solutions to an ODE. It includes a constant of integration, which accounts for any arbitrary values that may be present in the equation.

4. How do you find the constant of integration?

To find the constant of integration, you need to use initial conditions or boundary conditions. These are known values of the function or its derivative at a specific point, which can be used to determine the value of the constant.

5. Can an ODE have multiple solutions?

Yes, an ODE can have multiple solutions. This is because the general solution includes a constant of integration, which can take on different values and result in different solutions. However, initial or boundary conditions can help narrow down the solution to a specific one.

Suggested for: Solving ODE to find general solution

Replies
3
Views
545
Replies
3
Views
119
Replies
16
Views
380
Replies
1
Views
632
Replies
3
Views
400
Replies
4
Views
345
Replies
1
Views
434
Back
Top