# Math Help - DEs and projectiles

1. ## 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.

2. 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?

3. ## 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.....

4. 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 $a=-g$)

5. ## Thanks

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