Kinematics: Angled Projectile Launch

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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
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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
 

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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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