How Far Does the Arrow Fall Short of the Target?

Click For Summary
SUMMARY

The problem involves calculating how far an arrow falls short of a target when fired horizontally at a speed of 89 m/s from a height of 1 m. The arrow travels a horizontal distance of 60 m before hitting the ground. The solution reveals that the arrow strikes the ground 20 m short of the target. Key equations used include D = vt for horizontal motion and D = Vit + (1/2)at² for vertical motion, with vertical acceleration set at -9.8 m/s².

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Knowledge of horizontal and vertical motion components
  • Ability to interpret and apply positive and negative signs in physics
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to set up motion tables for horizontal and vertical components
  • Explore the effects of initial height on projectile trajectories
  • Practice solving similar problems involving horizontal projectile motion
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for problem-solving strategies in teaching these concepts.

chrozer
Messages
6
Reaction score
0

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 = Vit + (1/2)at2
Vf = Vi + at
Vf2 = Vi2 + 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.
 
Last edited:
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
12
Views
3K
Replies
14
Views
3K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K