Ballistics - Find velocity of arrow

[msadegian]

Homework Statement

You are a hunter who enjkoys shooting wild game with a bow. A moose, 100.0m away is charging with a velocity of 15 m/s toward the hunter. The hunter fires the bow at an angle of 30.0 degrees from the ground. What velocity must the arrow be fired at in order to hit the running moose?

I've tried and tried with this question, but it's so hard. My teacher said he got it from a student of his who went to NAIT (a tech institute) and we are only in Gr.11. Any help would be appreciated!

Homework Equations

d=vt
d=1/2gt²

The Attempt at a Solution

no clue at all!

 

[djeitnstine]

Only the horizontal velocity matters. And what parameter will be the same when the arrow and the moose meets?

 

[Delphi51]

A fine problem!
In grade 11, you should have the standard way of solving a trajectory problem.
You resolve the initial velocity into its horizontal and vertical components.
Then you make two headings for horizontal and vertical.
For horizontal, you write d = vt, using your horizontal initial velocity.
For vertical, you write v = vi + at and d = vi*t + .5a*t^2 using your vertical initial velocity.
This time you can't solve this because the initial velocity is unknown.
You must use the fact that the horizontal distance traveled by the arrow plus the distance traveled by the moose adds up to 100m. This gives you an equation relating time and v. A second time and v equation can be found from the vertical part by setting the vertical distance equal to zero (assuming that the arrow falls to its initial height when it reaches the moose). Solve these as a system of equations to find the velocity and time (between 2 and 3 seconds!).

A friend of mine had this experience and managed to get an arrow into the moose before it hit him!

 

[msadegian]

yes, thank you very much for you help. However i am still confused and unsure of how to do this.

 

[myxomatosii]

Thats a fast moose, personally I'd hide in a tree til it passes and go home.

 

[alexas]

A fine problem!
In grade 11, you should have the standard way of solving a trajectory problem.
You resolve the initial velocity into its horizontal and vertical components.
Then you make two headings for horizontal and vertical.
For horizontal, you write d = vt, using your horizontal initial velocity.
For vertical, you write v = vi + at and d = vi*t + .5a*t^2 using your vertical initial velocity.
This time you can't solve this because the initial velocity is unknown.
You must use the fact that the horizontal distance traveled by the arrow plus the distance traveled by the moose adds up to 100m. This gives you an equation relating time and v. A second time and v equation can be found from the vertical part by setting the vertical distance equal to zero (assuming that the arrow falls to its initial height when it reaches the moose). Solve these as a system of equations to find the velocity and time (between 2 and 3 seconds!).

A friend of mine had this experience and managed to get an arrow into the moose before it hit him!

Did he do all these calculations on the fly in his head as the moose was coming at him? If so that must have been intense! haha
I'm glad your friend pulled it off, i really doubt i could solve a physics problem like that even if my life depended on it.

 

[msadegian]

A fine problem!

You must use the fact that the horizontal distance traveled by the arrow plus the distance traveled by the moose adds up to 100m. This gives you an equation relating time and v. A second time and v equation can be found from the vertical part by setting the vertical distance equal to zero (assuming that the arrow falls to its initial height when it reaches the moose).

Okay I Think I sort of understand what you are trying to say. If you put d=vt are you supposed to put in 100=15*t and solve or just set t=d/15 ?
and when you put 0 in for your vertical component i understand that but you get 0=vi(d/15)+.5(9.81)(d/15)² then what are you supposed to solve for??

Thanks so much for your help!I appreciate it!

 

[Delphi51]

In grade 11, you should have the standard way of solving a trajectory problem.
You resolve the initial velocity into its horizontal and vertical components.

I leave the vertical component to you.

Then you make two headings for horizontal and vertical.
For horizontal, you write d = vt, using your horizontal initial velocity.

HORIZONTAL
x = v*cos(30)*t
I leave the vertical part to you.

I believe the vertical part is important, djeitnstein. Shooting at 30 degrees above horizontal, you could easily shoot over the moose's head.
My friend shot from MUCH closer with an angle of zero.
He was very lucky and very scared!
The moose was badly injured but ran away and it took several hours to find him and put him out of his misery.

 

[msadegian]

I'm confused as to where the 100m or 15m/s comes in??
and once you have the horizontal equation and vertical equation what do you do?do you need to solve for time first? I'm sorry it's just we were never taught this so I am pretty much clueless!

 

[Delphi51]

You must use the fact that the horizontal distance traveled by the arrow plus the distance traveled by the moose adds up to 100m. This gives you an equation relating time and v.

distance traveled by arrow + distance traveled by moose = 100
v*cos(30)*t + 15*t = 100
Yes, this is a bit beyond grade 11 physics in the mathematical complexity, but the concepts are within reach. You must show some of your work on the vertical part before I can help any more.

 

[msadegian]

is it something like 0=vsin30*t+0.5(9.81)t²?
or do you put vsin30 into the v=vi+at and get v=vsin30+9.81t?

i know i should be able to do this and when I realize how to do it I'll probably feel pretty dumb for not understanding!

 

[Delphi51]

0=vsin30*t+0.5(9.81)t2?
or do you put vsin30 into the v=vi+at and get v=vsin30+9.81t?

You need a minus sign on the 9.81 in both of those. The second one is easier to solve than the first if you realize that the final velocity will be -v*sin(30). Remember the old adage, "what goes up will come back down at the same speed".

Alternatively, you can temporarily take t to be the time to maximum height so Vf = 0 in
Vf = v*sin(30) - 9.81t. The t you find must then be doubled to get the time of flight.