# What is the velocity of a dart that is fired?

1. Oct 22, 2015

### DeathEater

1. The problem statement, all variables and given/known data
a hunter is trying to kill an armadillo that is 70 m away. To kill it, the hunter fires a dart at an angle 30 degrees above the horizontal. What must be the velocity of the dart in order to hit the armadillo?

2. Relevant equations
Vx=Vcosθ
Vy=Vsinθ
Δx= (Vix)(time)+½(acceleration)(time)2

3. The attempt at a solution
I assume that this will take on a shape of basically two right triangles back to back, since that will mimic the shape of the parabola the projectile makes. I broke the angle into components (only for one of the right triangles, and to find the distance traveled in the y direction and the hypotenuse, which is the dart basically) and I got that the hypotenuse is cos (30)= 35/ hypotenuse (I used 35 since it is only one of the triangles, so the base is half the distance traveled in the x direction). After solving I got the hypotenuse = 40.4145 m and using the same method and Tangent, got that the y-component is 20.207 m for one of the triangles.

I do not really know where to go after breaking it into components though, and if I made mistakes above, I would like to know so that I may correct and try to understand better

2. Oct 22, 2015

### Simon Bridge

Bad approach... do not approximate a parabola by two triangles... its way way off.
Note: you do not need the tangent or the hypotenuse to get components...

Try using kinematics or Newton's laws instead.
Velocity-time graphs can help a lot here: you will wantbto bresk the initual velocity into components.

Last edited: Oct 22, 2015
3. Oct 22, 2015

### Mister T

Because the acceleration in the x-direction is zero, this last equation should instead be

$\Delta x=(v_{ix})(\mathrm {time})$

But you will need an additional equation that includes $v_{iy}$ and involves the acceleration in the y-direction.

4. Oct 22, 2015

### DeathEater

my teacher has not taught this before, we have always had a given velocity, so it was very confusing when there wasn't one. And my teacher was the one who drew the triangle... I really do not know what to do. I am appreciative of all help and I really want to understand this.

5. Oct 22, 2015

### Mister T

Well, you can draw triangles to illustrate $\vec{v}$ and it's components, but not to approximate the the shape of the projectile's path.

Do you have a textbook for this course? Try looking up "range of a projectile".

6. Oct 22, 2015

### DeathEater

this question was on a previous test. He didn't explain it, and the things we learned prior to taking the test was kinematic equations and how to use SOHCAHTOA to break things into components. Using that knowledge only (and no we do not have a textbook) how could I solve this? I know that the point of these forums is to try to prick and prod until I come up with the discovery on my own, but I really have absolutely no idea what to do. We haven't discussed Newtons laws or anything either

7. Oct 22, 2015

### Mister T

You need to take each kinematic equation for one-dimensional motion and turn it into a pair of equations, one for the horizontal component of the motion and one for the vertical component.

I suggest a trip to the library or bookstore to find an introductory physics textbook.

See Post #3.

8. Oct 22, 2015

### DeathEater

well how would I be able to use the equations together to find it?

9. Oct 22, 2015

### Mister T

The are connected by the time t. The time it takes for the horizontal position to change equals the time it takes for the vertical position to change.

10. Oct 22, 2015

### DeathEater

so I use 70 m for the x, and then using the angle and x do I find the y component which is 20.207 m?(I think)

11. Oct 22, 2015

### Mister T

$\Delta x=70 \ \mathrm{m}$, that was given.

But $\Delta y \neq 20.207 \ \mathrm{m}$ as we've already told you.

12. Oct 22, 2015

### DeathEater

I do not know how to go about solving it and how to find the change in y as I have already said many times before.

13. Oct 22, 2015

### Mister T

We know. Sorry, but if you're not willing to follow our instructions there's no way we can help you learn.

14. Oct 22, 2015

### DeathEater

I am following the instructions, you are just assuming that I know how to do things that I do not and have not been taught

15. Oct 22, 2015

### Mister T

You've been to the library or bookstore, picked up an introductory physics textbook, taken each one-dimensional kinematics equation and turned it into a pair of equations, and looked up "range of a projectile"?

I'm not assuming you know how to do each of those things. I am assuming you can do the first two. And I'm hoping you can learn the others. Because I know that if you get stuck along the way we are here to help you.

16. Oct 22, 2015

### DeathEater

well I did look up range of a projectile and got the equation . The only problem is that if I am doing this for X, then acceleration from gravity (g) would be zero...

17. Oct 23, 2015

### Simon Bridge

... back up.
Is this part of a physics course?
Have you completed prior material on kinematics?
Have you completed prior material on Newton's laws of motion?
Have you completed prior material on vectors?

... without these topics, you cannot complete the assignment.
Physics courses always complete these topics before starting on projectile motion, which is why knowledge of these topics is being assumed. If you do not have knowledge of these topics, you will have to go back over your course and back tgrough your text book and notes to gain the needed knowledge. We cannot just give you physics lessons here.

Last edited: Oct 23, 2015
18. Oct 23, 2015

### Simon Bridge

Note... g appears in the equation for range because $v_0$ is the magnitude of the initial velocity not just the horizontal velocity. You need both horizontal and vertical components to complete the assignment because gravity affects the motion of the dart... so gravity will appear in its equations of motion.

You can use the equation to complete the assignment, but you won't understand it.

19. Oct 23, 2015

### DeathEater

yes, it is part of a physics course, no we have not learned about the laws of Newton, and no we do not have a textbook and the notes are minimal. This problem was on a previous test of mine, so my teacher assumed that I should know how to do this, which I don't.

20. Oct 23, 2015

### DeathEater

using the equation though, would it be 70= ((vo2)/ 9.81)* sin (2*30) ----> 70 /sin(60)= (vo2)/9.81 ----> 80.83= (vo2)/ 9.81 -----> 80.83*9.81= vo2 ----> 792.12=vo2------> vo =28 m/s (after finding the square root)