Finding initial velocity and time Please

[red_viper_88]

Finding initial velocity and time! Please please help!

1. A missile is aimed to hit a target 650 km away. If the missile is fired at an angle of 20 degrees with the horizontal, what should be its launch speed so that it will hit the target? Once fired, how long does the missile take to hit the target?

Homework Equations

I'm not certain which equation to use to get the initial velocity. But I think I may have to use the 3rd equation of motion: velocity final = velocity initial + 2 (acceleration)(displacement)...however, I'm missing a lot of the variables. So i don't know where to even get started! And then for the second part, we need to find time! So I think maybe using the 1st equation of motion or the 2nd equation of motion? velocity final = velocity initial + (acceleration) (time)?

The Attempt at a Solution

As I said, I'm not even certain where to begin! I am given the information that the direction is 20 degrees and that the displacement of the hypotenuse is 650 km or 650 x 10^3. Then I found the opposite is 222 km and the adjacent is 611 km by using the law of sin and cos. I really think I'm way of track, and I need some assistance getting back on track please!

 

[Cyosis]

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.

 

[cristo]

But I think I may have to use the 3rd equation of motion: velocity final = velocity initial + 2 (acceleration)(displacement)

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

 

[red_viper_88]

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.

x component: velocity of x = v cos Ɵ
y component: velocity of y = v sin Ɵ

what do i do next?

 

[red_viper_88]

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

yes! that's the one! do i use that one?

 

[Cyosis]

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

 

[red_viper_88]

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

I believe there is an accleration in the y-direction, i don't know what it is and how to find it. And acceleration in this case would be constant. I don't think there is an acceleration in the x-direction. Just a velocity.

I do not know the "kinematic equations for constant acceleration and constant speed".

 

[Cyosis]

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:
s=\frac{1}{2} a t^2+v_0 t+x_0

Constant speed:
s=v t

 

[red_viper_88]

gravity?

so I'm substituting -9.8 for acceleration (g)?

what is s in the equations u listed? displacement?

and i don't know what time is...so I'm kind of confused as to how to use those equations

 

[Cyosis]

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.

 

[red_viper_88]

displacement = uniform velocity x time
this uniform velocity = displacement / time
constant velocity = 650 x 10^3 m / time
??
i'm very lost!
there is another equation i found is v = at...but I'm still messed up with the t variable! i can't substitute anything for that in order to find velocity. I'm sooooooooooo lost!

 

[Cyosis]

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

For the y-direction this yields:
y=-\frac{1}{2}g t^2+v_{y,0}t
At the impact site the height is 0 and v_y0=v sin 20. Let's enter this into the equation, which yields:

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

For the x-direction we have x=v_{x,0}t. 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:

650*10^3=v \cos(20)t. 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.

<br /> \begin{align}<br /> 650*10^3 &amp; =v \cos(20)t<br /> \\<br /> 0 &amp; =-\frac{1}{2}g t^2+v \sin(20)t<br /> \end{align}.

 

[red_viper_88]

how is it solvable with 2 unknowns?

 

[Cyosis]

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

 

[red_viper_88]

i don't understand how i combine them

 

[Cyosis]

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

 

[red_viper_88]

so for the first one t = (650 * 10^3) / v cos (20) ?

 

[Cyosis]

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

 

[red_viper_88]

i can't seem to get v alone!
= -4.9 (650 x 10^5 / 0.88 v^2) + 0.34 v (650 x 10^3 / 0.34 v)

i simplified the second equation a lil more, so i can work with it easier before substituting

 

[Cyosis]

You're making it too hard on yourself.

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

 

[red_viper_88]

i got 1/2g = v sin(20) / t...is that right?

thus i would get 4.9 = v sin 20 / t

and then i would substitute that equation i made for t?

and get 4.9 on one side and 0.34v / (650 x 10^3 / 0.94v) on the other side?

 

[Cyosis]

That's correct so far.

 

[red_viper_88]

oh wow! i really like the formula u gave me better! let me try that!

so 4.9 (650 x 10^3 / 0.94v) = 0.34v
then 4.9 (650 x 10^3) = 0.34v x 0.94 v
then 4.9 (650 x 10^3) = 0.34 x 0.94 v^2
then v^2 = 4.9 x (650 x 10^3) / 0.34 x 0.94
then v = 3156.83?
that is one big number!

 

[Cyosis]

It's a big number yes. Why are you at a dead end now? Look at your equations and see what information you have. You have everything except time.

 

[red_viper_88]

is it the correct number?

 

[Cyosis]

Yes I got the same number.

 

[red_viper_88]

i had a friend help me finish it up, can u please double check the answer! thanks!
vertical component = v sin 20
3km sin 20
s = ut + 1/2at^2
s = 0
u = 3
a = -9.8
thus, 4.9t^2 - 1076t = 0
t(4.9t - 1076) = 0
4.9t = 1076
t = 1076 / 4.9 = 219 seconds

 

[Cyosis]

Correct, but again you make it harder than you have to. You had v so why not use x=vt->t=x/v. Always pick the easiest equation to minimize the chance of errors.

 

[red_viper_88]

yup! u r right!
so the answer of 219 secs matches urs?
or is it 220 secs?

or with the formula u told me to use, would be:
t = (650 x 10^3) / 3157 = 206 secs

can u please double check that?
thanks!

 

[Cyosis]

I haven't looked at the decimal digits. If you want to get the most accurate result don't round until the final answer. It may be the difference between 219.4 for example and 219.5=220.

 

[Cyosis]

or with the formula u told me to use, would be:
t = (650 x 10^3) / 3157 = 206 secs

Remember x=v_x t. You entered the total speed, but instead you need to enter v_x=v cos(20). That's why you get a different answer.

 

[red_viper_88]

so which is the correct way to do it?
and what would u advise me to put as my final answer?

 

[Cyosis]

Picking either equations will do, however always pick the easiest to minimize the errors. Secondly for optimal accuracy don't calculate the cos/sin then round them and then use them in further calculation. Just keep them as cos/sin. Same for the other variables if you do that you get the following answers.

v=3149.61 m/s
t=219.619 s

You see that rounding constantly during the calculation introduces errors.

 

[red_viper_88]

thank you so much for all your time, help, and patience![/color]