# DEs and projectiles

• Sep 20th 2009, 06:33 PM
latavee
DEs and projectiles
This problem relates to projectiles without air resistance, but using DEs Solve the IVP to find an expression for v(t) in terms of g and t(no m dependence)

1st: Given mv'(t)=-mg ; v(0)=v0

2nd: v'(t)=-g (dividing by m, since there is no mass dependence)
3rd: integrate v(t)=-9.8t

I'm stuck here, because I am to use the result from v(t) and solve for the IVP: x'(t)=v(t) ; x(0)=0 (height above ground of the projectile) in terms of v0,g, and t.
• Sep 20th 2009, 06:40 PM
VonNemo19
Quote:

Originally Posted by latavee
This problem relates to projectiles without air resistance, but using DEs Solve the IVP to find an expression for v(t) in terms of g and t(no m dependence)

1st: Given mv'(t)=-mg ; v(0)=v0

2nd: v'(t)=-g (dividing by m, since there is no mass dependence)
3rd: integrate v(t)=-9.8t

I'm stuck here, because I am to use the result from v(t) and solve for the IVP: x'(t)=v(t) ; x(0)=0 (height above ground of the projectile) in terms of v0,g, and t.

Could you post the original problem, please?
• Sep 20th 2009, 06:51 PM
latavee
problem detail
A projectile is shut up vertically without taking air resistance into account, considering a basic vertical throw.After being shot, the only force acting upon the projectile is gravity -mg, with negative sign indicating gravity act downwards. Thus Newton's Law gives ma(t)=-mg. With a(t)=v'(t) and use v(0)=vo(not) as initial condition this gives the first order initial value problem for velocity:

mv'(t)=-mg, v(0)=vo....my original post goes from here.....
• Sep 20th 2009, 07:46 PM
Chris L T521
Quote:

Originally Posted by latavee
A projectile is shut up vertically without taking air resistance into account, considering a basic vertical throw.After being shot, the only force acting upon the projectile is gravity -mg, with negative sign indicating gravity act downwards. Thus Newton's Law gives ma(t)=-mg. With a(t)=v'(t) and use v(0)=vo(not) as initial condition this gives the first order initial value problem for velocity:

mv'(t)=-mg, v(0)=vo....my original post goes from here.....

First integrate both sides to get $v(t)=-gt+c$.

Applying the initial condition $v(0)=v_0$, we have $v(0)=v_0=-g(0)+c\implies c=v_0$.

Thus, $v(t)=-gt+v_0$.

Since $x^{\prime}(t)=v(t)$, it follows that if we integrate both sides of the equation, we have

$x(t)=-\tfrac{1}{2}gt^2+v_0t+c$.

Now, supposing that $x(0)=x_0$, we have $x(0)=x_0=-\tfrac{1}{2}g(0)^2+v_0(0)+c\implies c=x_0$.

Thus, $x(t)=x_0+v_0t-\tfrac{1}{2}gt^2$.

In your first post, I saw that you let $x_0=0$. So it turns out that the equation you're interested in is $x(t)=v_0t-\tfrac{1}{2}gt^2$.

(If you're familiar with physics, we take note that this is the kinematics equation $s_f=s_0+v_0t+\tfrac{1}{2}at^2$, where we have $s_0=0$ and $a=-g$)
• Sep 20th 2009, 08:14 PM
latavee
Thanks
Ah! Thank you...you made that so understandable!