# Solving ODE to find general solution

by andrey21
Tags: solution, solving
 P: 466 [b]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. 2. Relevant equations y is the distance above the ground of an object y' is the vertical velocity y'' is vertical accelaration 3. 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
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P: 6,202
 Quote by Jamiey1988 3. 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.
 P: 466 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?
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P: 6,202
Solving ODE to find general solution

 Quote by Jamiey1988 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.
 P: 466 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?
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P: 6,202
 Quote by Jamiey1988 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.
 P: 466 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?
 HW Helper P: 6,202 Yes that looks correct.
 P: 466 Thanks for all your help
 P: 466 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.

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