How far does the object travel horizontally? Two-Dimensional Motion

  • Thread starter Thread starter eioz
  • Start date Start date
  • Tags Tags
    Motion Travel
Click For Summary

Homework Help Overview

The problem involves a rifle bullet fired from a cliff at an angle, requiring the calculation of horizontal distance traveled. The context is two-dimensional motion, specifically projectile motion, with given parameters including initial velocity, angle, and height of the cliff.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial calculations of vertical and horizontal components of velocity. Questions arise regarding the use of equations to find time and how vertical travel relates to horizontal distance.

Discussion Status

The discussion is ongoing, with participants exploring different methods to calculate the time of flight based on vertical motion. Some guidance has been offered regarding the relationship between vertical distance, initial velocity, and time, but no consensus on a complete approach has been reached.

Contextual Notes

Participants note the challenge of working independently without direct instructional support, which may affect their confidence in applying the equations correctly.

eioz
Messages
2
Reaction score
0

Homework Statement


A rifle bullet is fired from the top of a cliff at an angle of 30° below the horizontal. The initial velocity of the bullet is 800 m/s. If the cliff is 80 m, high, how far does it travel horizontally?

θ = 30°
v0 = 800 m/s
y = 80 m


Homework Equations


I may be totally off here with these equations (I'm part of an independent study program and their physics curriculum doesn't even have an online teacher, so I've been flying blind since day one), but I think these are relevant:

vy = Vsinθ
|v| = √vx2 + vy2

The Attempt at a Solution


vy = 800sin30 = 400

800 - 400 = 400 = vx

I then thought I could go from there to solve for x using x = x0 + v0xt + 1/2at2, but I don't have anything for t, so I'm stuck.
 
Physics news on Phys.org
With 800 - 400 = 400 = vx you got the right answer, but only accidentally. To get the correct answer the correct way, you must use the second equation you listed in the relevant eq-s section.

Before you you compute the horizontal distance, ask yourself how far the bullet will travel vertically. And how long that will take.
 
I actually was using the equation I listed in the relevant eq-s section, but I do see how that isn't entirely clear.

I unfortunately really do not know where to go from where I'm at. The bullet will travel 80 m vertically, yes? But I don't understand how to use any of that to help me get time.

Could I use y = y0 + v0t to find out time, or does that not work here?
 
Last edited:
eioz said:
The bullet will travel 80 m vertically, yes? But I don't understand how to use any of that to help me get time.

The distance of vertical travel depends on the initial vertical velocity, acceleration and time. You know all of these except the time, so you should be able to find it.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K