Kinematics: Angled Projectile Launch

AI Thread Summary
The discussion focuses on determining the initial velocity of a projectile launched from a cannon at a 20° angle. Participants analyze the problem by breaking it into x and y components and applying the kinematic equation d = (Vi)(t) + 1/2(a)(t)^2. Confusion arises regarding the initial vertical velocity (v0y), with suggestions to correct algebraic mistakes and re-evaluate calculations. One participant recalculates and arrives at an initial velocity of 44.5 m/s but seeks further verification on their previous calculations. Overall, the thread emphasizes collaboration and learning in solving kinematics problems.
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: 443
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top