Free Fall Problem: Calculating Projectile Motion from a Cliff Edge

  • Thread starter Thread starter shawonna23
  • Start date Start date
  • Tags Tags
    Fall Free fall
Click For Summary

Homework Help Overview

The discussion revolves around a free fall problem involving a pellet gun fired from a cliff. The scenario includes calculating the height the pellet would reach if fired upward, given its downward speed upon impact.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and energy conservation principles. Questions arise regarding the assumptions made about the pellet's speed and whether it represents terminal velocity.

Discussion Status

There is an ongoing exploration of different approaches, with some participants suggesting the use of energy conservation to find the initial velocity. The discussion reflects a mix of interpretations and attempts to clarify the correct equations to use.

Contextual Notes

Participants are navigating assumptions about terminal velocity and the appropriateness of the equations being applied in the context of the problem.

shawonna23
Messages
146
Reaction score
0
Help! Free Fall Problem

A pellet gun is fired straight downward from the edge of a cliff that is 13 m above the ground. The pellet strikes the ground with a speed of 30 m/s. How far above the cliff edge would the pellet have gone had the pellet been fired straight upward?
 
Physics news on Phys.org
What have you done so far?
 
I started using the equation :

y=-1/2 gt^2-30t+0
 
are you assuming that 30 m/s is not the pellets terminal velocity?
 
Have you learned about energy conservation yet? If so then you can use it to find what the "muzzle velocity" is. Basically it says that [itex]\frac{1}{2}mv^2_{final} - mg \times 30 = \frac{1}{2}mv^2_{initial}[/itex], i.e. potential + kinetic energy is constant.
 
yes, I'm assuming that 30m/s is not the pellets terminal velocity
 
shawonna23 said:
I started using the equation :

y=-1/2 gt^2-30t+0

Hmmm, that's not the right equation - did you understand the remarks Tide made about energy conservation?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
7K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
5K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
Replies
7
Views
3K
Replies
13
Views
4K