A question and no idea what to do

  • Thread starter Thread starter ouse
  • Start date Start date
  • Tags Tags
    Idea No idea
Click For Summary

Homework Help Overview

The discussion revolves around determining the initial velocity of a projectile launched horizontally, which falls a vertical distance of 1.5 meters while traveling horizontally 16 meters. Participants are exploring the kinematics involved in projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss dividing the motion into horizontal and vertical components, questioning the applicability of this method without angles. There is an exploration of the relevant equations of motion and their derivations.

Discussion Status

The discussion has progressed with some participants expressing understanding of the equations provided. There is an acknowledgment of the assumptions made regarding initial vertical velocity and the effects of gravity. However, there remains a request for further assistance on related questions, indicating ongoing exploration.

Contextual Notes

Some participants express concerns about insufficient information and the nature of the problem, particularly regarding the lack of angles for determining components of motion.

ouse
Messages
33
Reaction score
0
determine the inital velocity of a projectile that is luncked horizantaly and falls 1.5m while moving 16m horizantaly
i tried everything all i figerd out is that they r displacmednts the accelaration is 9.8 and nothing i don't have enogh info to do anything
please help me
 
Physics news on Phys.org
Divide the motion into x and y components and deal with them as if you were dealing with one-dimensional motion.
 
can u please elaborate
 
and any way that doesn't work becuse i have no angles to detrime wat the x and y components are
 
that would only work if these were vectors not displacment
 
Well...
x = v_0t
0 = y_0 -\frac{1}{2}gt^2
Does that help at all?
 
yeah it helped i acctualy figerd the question thanks sooo much
 
now can u please look at my other thread and please ohh please help with question 1 b,c and q 2
 
ouse said:
yeah it helped i acctualy figerd the question thanks sooo much
Glad to hear that :).

Did you understand where the equations I gave came from?
 
  • #10
ummm well yeah for one of them
 
  • #11
u assumed viy is 0 and took it our of the equation right
 
  • #12
The full equations:
x = x_0 + v_{x0}t + \frac{1}{2}a_xt^2
y = y_0 + v_{y0}t + \frac{1}{2}a_yt^2
Let x_0 = 0. As the projectile's fired horizontally: v_{y0} = 0 thus v_{x0} = v_0. No force affects the projectile in x-direction, so: a_x = 0. In y-direction: a_y = -g (- as it's downward). The final altitude of the projectile is y = 0.

Hence
x = v_0t
0 = y_0 -\frac{1}{2}gt^2
 
  • #13
i was woundring if u could help me with the other thread please help me i really need it i have a test i need to pass
 
  • #14
thx for the explanation
 

Similar threads

Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
7
Views
9K
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K