Kinematics: Angled Projectile Launch

Click For Summary

Homework Help Overview

The discussion revolves around determining the initial velocity of a projectile launched from a cannon at a 20° angle. The problem involves kinematics and the decomposition of motion into horizontal and vertical components.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting the motion into x and y components and using kinematic equations. There are questions about the assumptions made regarding initial vertical velocity and algebraic manipulations. Some participants express confusion about the variables involved and seek verification of their calculations.

Discussion Status

The discussion is active, with participants providing feedback on each other's approaches and calculations. There is an acknowledgment of mistakes and attempts to clarify the relationships between the components of velocity. Some guidance has been offered regarding the algebra involved and the use of trigonometric relationships.

Contextual Notes

Participants mention feeling uncertain about their understanding and calculations, indicating a learning environment where assumptions and methods are being critically examined. There is a sense of camaraderie among those who identify as beginners in physics.

SlooM
Messages
3
Reaction score
0

Homework Statement


Determine the initial velocity of the projectile as it comes out of a cannon at a 20° angle.


Homework Equations


d = (Vi)(t) + 1/2(a)(t)^2


The Attempt at a Solution


I've split the question into both x and y components and solved for both using the equation above. Although I'm honestly unsure if I'm doing this correctly, would like some verification on my answer and point out any mistakes. Much appreciated! (Image below)

http://imgur.com/pPWO2jG
 

Attachments

  • Scan001.jpg
    Scan001.jpg
    26.7 KB · Views: 465
Last edited:
Physics news on Phys.org
Hi SlooM,

Are you sure v0y is 0 in the first equation that you've used to find t?
 
Ah, I think that may have been my mistake which is kind of why I look at it and it doesn't make sense. I just put 0 for v0y and assumed it canceled that half of the equation. What would I put though, I feel like I don't have enough variables and am super confused.
 
In addition to what Sunil has stated, there was also an algebra mistake at the end preventing you from getting the correct answer.

\cos\theta = \frac{v_x}{v_i}

to

(cos\theta)v_x = v_i

instead of

v_i =\frac{v_x}{cos\theta}


An easier way to do this problem it to remember that your x-component of velocity already has a cosine term in it. So once you solve for time you can simply say

x(t) = V_otcos\theta
 
Last edited:
Oh! I can't believe I didn't see that. So I redid the algebra at the end and got 44.5 m/s but are my calculations in the previous x and y calculations correct? I keep second guessing myself :/

I'm a rookie at physics but trying to improve. I appreciate all of your help and it means a lot.
 
SlooM said:
Oh! I can't believe I didn't see that. So I redid the algebra at the end and got 44.5 m/s but are my calculations in the previous x and y calculations correct? I keep second guessing myself :/

I'm a rookie at physics but trying to improve. I appreciate all of your help and it means a lot.

Hey we are all rookies :smile:

We get two equations right?

-1.5= v_{0y}t + \frac{g}{2}t^2
23=v_{0x}t

All we have to do is eliminate t. Moreover, v0x and v0y are related. So, you should be able to find v0x explicitly (and hence v0y).

Could you check you answer again? I seem to be getting something different.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
4K
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 15 ·
Replies
15
Views
27K
  • · Replies 15 ·
Replies
15
Views
10K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
27
Views
3K
Replies
3
Views
2K