Finding initial velocity and time Please please help

Re: Finding initial velocity and time!!! Please please help!!!

I don't think they mean that the missile continues to fly in a line with an angle of 20 degrees with the horizontal. This would mean it would hit some aerial target.

The target is just 650km away at the same height as the launch spot. After the launch the missile will follow a parabolic trajectory. Split the velocity up in an x and y component. Then write down an equation that gives the height of the missile as a function of time and another equation that gives the horizontal displacement as a function of time. Then solve.

 

Re: Finding initial velocity and time!!! Please please help!!!

This equation isn't correct. The one you are thinking of is [itex]v^2=u^2+2as[/itex].

 

Re: Finding initial velocity and time!!! Please please help!!!

Is there an acceleration in the y-direction, if so what is it and is it constant? Is there an acceleration in the x-direction, if so what is it and is it constant?

You know the kinematic equations for constant acceleration and constant speed right? Find out which one applies to the y-direction and which one applies to the x-direction

 

Re: Finding initial velocity and time!!! Please please help!!!

You're correct in both cases. Finding the acceleration in the y-direction isn't very hard though. What is the only force acting on the missile after it has been launched?

I am quite certain you do know the kinematic equations for constant acceleration and constant velocity.

Constant acceleration:
[tex]s=\frac{1}{2} a t^2+v_0 t+x_0[/tex]

Constant speed:
[tex] s=v t[/tex]

 

Re: Finding initial velocity and time!!! Please please help!!!

Yes s is displacement and you're correct about the acceleration.

For the y-direction you know:
There is an acceleration of -g in the y direction.
You know v_y a a function of the angle
Now all you need to know is what the height is at the impact.

Then you fill all the data into the first equation.

For the x-direction you know:
There is no acceleration so v_x is v_xinitial.
You know v_x as a function of the angle.
You know the horizontal displacement.

Now fill the data into the equation for constant velocity.

 

Re: Finding initial velocity and time!!! Please please help!!!

Don't worry too much about the time and lets call the vertical displacement y and the horizontal displacement x from now on.

For the y-direction this yields:
[tex]y=-\frac{1}{2}g t^2+v_{y,0}t[/tex]
At the impact site the height is 0 and v_y0=v sin 20. Lets enter this into the equation, which yields:

[tex]0=-\frac{1}{2}g t^2+v \sin(20)t[/tex].
Unknowns t and v. One equation two variables so this isn't solvable. Luckily we have more information!

For the x-direction we have [itex]x=v_{x,0}t[/itex]. We know that x is 650km at the impact site and we also know that v_x0=v cos(20). Entering this information into the equation gives us:

[tex]650*10^3=v \cos(20)t[/tex]. Again one equation and two unknown variables, however these are the same two unknown variables as in the previous equation.

Therefore we have a set of two equations and two variables, which is solvable.

[tex]
\begin{align}
650*10^3 & =v \cos(20)t
\\
0 & =-\frac{1}{2}g t^2+v \sin(20)t
\end{align}[/tex].

 

Re: Finding initial velocity and time!!! Please please help!!!

You have two equations. Calculate t of impact as a function of v and plug that into the other equation.

 

Re: Finding initial velocity and time!!! Please please help!!!

Pick one of them and solve it for t. Show me the outcome.

 

Re: Finding initial velocity and time!!! Please please help!!!

Yes, so now you have found t as a function of v. Plug this t into the other equation and show me the result.

 

Re: Finding initial velocity and time!!! Please please help!!!

You're making it too hard on yourself.

We want to plug t into [itex]0=-\frac{1}{2}g t^2+v \sin(20)t[/itex]. We know that t is not 0. So we can divide by t without problems. This yields [itex]0=-\frac{1}{2}g t+v \sin(20)[/itex]. Now we bring the t part to the other side yielding [itex]\frac{1}{2}g t= v \sin(20)[/itex]. Entering all the data is much easier now, try again!