How Far Does the Arrow Fall Short of the Target?

[chrozer]

Homework Statement

An arrow is fired with a horizontal spped of 89 m/s directly at a target 60 m away. When it is fired, the arrow is 1 m above the ground. How far short of the target is it when it strikes the ground?

Homework Equations

D = vt
D = V_it + (1/2)at²
V_f = V_i + at
V_f² = V_i² + 2ad

The Attempt at a Solution

Basically my teacher says to set up a table, one side horizontal values and the other vertical values.

The values I know for:

• Horizontal - velocity = 89, distance = 60
• Vertical - acceleration = -9.8, distance = 1

Then that is where I get stuck...help? BTw the anser is 20 m short, but my teacher wants me to show work.

 

[Redbelly98]

There are some more items to fill out in your table.

Horizontal: time?
Vertical: time?

Also, there are a couple of problems with the values you do have.

1. If the arrow falls short of the target, it does not go the full 60m distance.

2. Acceleration is -9.8 m/s^2, so you are using upward=positive, downward=negative. But then you say the vertical distance traveled is +1m upward. Gotta watch those +/- signs.

 

[thomate1]

I would approach this problem by first identifying the known values and variables, and then applying the relevant equations to solve for the unknown variable.

Known values:
- Horizontal velocity (Vi) = 89 m/s
- Horizontal distance (D) = 60 m
- Vertical acceleration (a) = -9.8 m/s^2
- Starting vertical distance (Vi) = 1 m

Unknown variable:
- Horizontal distance at impact (Df)

Using the equation D = vt, we can first solve for the time it takes for the arrow to reach the target:
60 m = 89 m/s * t
t = 0.674 s

Next, we can use the equation D = Vit + (1/2)at^2 to solve for the final vertical distance (Df):
Df = 1 m + (89 m/s * 0.674 s) + (1/2 * -9.8 m/s^2 * (0.674 s)^2)
Df = 1 m + 59.986 m - 2.323 m
Df = 59.663 m

Since the arrow started at a height of 1 m and landed at a final vertical distance of 59.663 m, it has traveled a total vertical distance of 58.663 m before hitting the ground. This means it fell 58.663 m in 0.674 s.

Using the equation Vf = Vi + at, we can solve for the final vertical velocity (Vf) at impact:
Vf = 89 m/s + (-9.8 m/s^2 * 0.674 s)
Vf = 82.3 m/s

Now, using the equation Vf2 = Vi2 + 2ad, we can solve for the final horizontal distance at impact (Df):
0 = (89 m/s)^2 + 2 * (-9.8 m/s^2) * Df
Df = -801.8 m / (-19.6 m/s^2)
Df = 40.9 m

Since the target is 60 m away, and the arrow has traveled a horizontal distance of 40.9 m before hitting the ground, it is 60 m - 40.9 m = 19.1 m short of the target.

Therefore, the final answer is that the arrow falls