DEs and projectiles

**latavee** · September 20th 2009, 05:33 PM

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.

**VonNemo19** · September 20th 2009, 05:40 PM

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?

**latavee** · September 20th 2009, 05:51 PM

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.....

**Chris L T521** · September 20th 2009, 06:46 PM

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 $a=-g$ )

**latavee** · September 20th 2009, 07:14 PM

Ah! Thank you...you made that so understandable!

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