# Solve Problem: Arrow Shot Horizontally Hitting Target 25m Away

• mystry4
In summary, to solve the problem of an arrow being shot horizontally at a speed of 100m/s and hitting a bullseye 25m away, you can use the equations dhorizontal = Vot and dvertical = 1/2at^2. Set dhorizontal= 25 m, a= -9.81 m/s^2, and Vo= 100 m/s. Solve for t to find the time it takes for the arrow to reach the target, then use that value to solve for dvertical to determine where it will hit the target.

#### mystry4

I need to make sure I am using the correct formulas (dvertical = 1/2at^2 solving for t^2 and dhorizontal = Vot) for the following problem:
arrow shot horizontally - speed of 100m/s - at bullseye 25 m away. How long til it reaches target and where will it hit target.
Thanks!

Yes, if the arrow is shot horizontally, those are the correct equations. Set dhorizontal= 25 m, a= -9.81 m/s2, and Vo= 100 m/s. Solve the dhorizontal equation for t (how long till it reaches the target) and then use that value to solve for dvertical (where will it hit the target).

To solve this problem, we can use the equations of motion to determine the time it takes for the arrow to reach the target and where it will hit the target.

First, let's define our variables:

- dvertical: vertical displacement (in this case, it is 0 since the arrow is shot horizontally)
- dhorizontal: horizontal displacement (25 m)
- a: acceleration due to gravity (9.8 m/s^2)
- t: time (what we are trying to find)
- Vo: initial velocity (100 m/s)

Using the equation dhorizontal = Vot, we can rearrange it to solve for t:

t = dhorizontal / Vo

Substituting our values, we get:

t = 25 m / 100 m/s = 0.25 s

So it will take 0.25 seconds for the arrow to reach the target.

To find where it will hit the target, we can use the equation dvertical = 1/2at^2 and solve for dvertical:

dvertical = 1/2 * 9.8 m/s^2 * (0.25 s)^2 = 0.30625 m

Therefore, the arrow will hit the target 0.30625 meters below the bullseye.

Just a note, we assumed that the arrow was shot at ground level and there is no air resistance. If we take into account the height at which the arrow is shot and air resistance, the calculations may be slightly different. But for this problem, these equations should give a reasonable estimate.

## 1. How do you solve the problem of an arrow shot horizontally hitting a target 25m away?

The first step is to gather all the necessary information, such as the initial velocity of the arrow, the acceleration due to gravity, and the distance to the target. Then, using the equation x = v0t, solve for the time it takes for the arrow to travel 25m. Finally, use the equation v = v0 + at to find the velocity of the arrow at the target.

## 2. What factors affect the accuracy of an arrow shot horizontally?

The initial velocity of the arrow, the angle at which it is shot, and air resistance are all factors that can affect the accuracy of an arrow shot horizontally. Wind and other external factors may also play a role.

## 3. How does the distance of the target impact the trajectory of the arrow?

The distance of the target will impact the trajectory of the arrow because it affects the time it takes for the arrow to reach the target. The farther the target, the longer the arrow will be in the air, and the more it will be affected by external factors such as wind and air resistance.

## 4. Is it possible to hit a target exactly 25m away with an arrow shot horizontally?

Yes, it is possible to hit a target exactly 25m away with an arrow shot horizontally. However, this would require precise initial velocity, angle, and environmental conditions, making it a difficult task to achieve consistently.

## 5. How can the trajectory of an arrow be adjusted to hit a target at a specific distance?

The trajectory of an arrow can be adjusted by changing the initial velocity or the angle at which it is shot. Additionally, adjusting environmental factors such as wind and air resistance can also impact the trajectory of the arrow. Practice and experimentation are key to achieving a specific distance target consistently.