How fast was the arrow shot?

Question

You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 61.0 m away, making a 2.00 degrees with the ground. How fast was the arrow shot?

Answer 1

EDIT*James, the final velocity of the arrow can NOT be 0! You can see this, since as you said yourself S =V t, and if V = 0 then the arrow didn't travel any ground, which we know it went 61 m!*

Oh, I love these type of questions.

We'll need to make a table to keep all of our information straight.

This is a 2-dimensional problem (up/down, left/right) and involves gravity.

In the table, if we know the value, we'll plug it in, otherwise we'll put in variables.

x | y
init. distance = 0 | h
final distance = 61 m | 0
init. velocity = vx | 0
fin. velocity = vx | vfy
acceleration = 0 | -9.8 m/s/s
time t | t

Other information - the arrow is making a 2.00º angle with the ground.

We are ignoring air resistance (I'm assuming), so the acceleration in the x direction is 0. The initial velocity in the y direction is zero since the archer aimed parallel to the ground. One last assumption we are making is that the point where the archer is standing is at the same level as where the arrow landed (for example, the ground is completely flat).

Usually, we use the information from the y direction to determine the time, and from there we can determine the arrow's initial speed.

Before we can use any of this, however, we need to figure out the intial height of the arrow, which will require clever use of trigonometry.

Imagine a right triangle. The arrow is one vertex of the triangle, the archer's feet is another, and the bow is at the triangle's height.

|_\
|___\
|_____\
|______\
|______2º_\ <---arrow

|----61 m------|

We can find the initial height of the arrow using the equation

h = tan(2º) * 61 = 2.130 m

That's great, because now we can use a kinematic equation to determine the time it took for the arrow to hit the ground.

y = 1/2 g * t^2 + viy * t + h

Plugging in everything

0 = 1/2 * (-9.8) * t^2 + 0 * t + 2.13

Sovling yields

t = 0.659 s

Since the arrow travels 61 m in 0.659 s, we get

vx = 92.5 m/s.

That was fun :D

Answer 2

Tan 2 degrees =h divided by 61. where h is the heigth originally shot from the bow.

Then h=1/2 a t squared, solve for t which is time it was in the air. where a= gravitational constant

and ground distance S=V t =61 Solve for V

That is the velocity of air average,

and average is equal to initial V + V final divided by 2,

So the final is Zero, and you can see that V leaving the arrow quiver is 2 times the average as calculated above.