Kinematics: Angled Projectile Launch

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SUMMARY

The discussion focuses on determining the initial velocity of a projectile launched from a cannon at a 20° angle. Participants utilized the kinematic equation d = (Vi)(t) + 1/2(a)(t)^2 to break down the problem into x and y components. Key corrections were made regarding the calculation of the initial vertical velocity (v0y) and the algebraic manipulation of the cosine relationship between the components. The final calculated initial velocity was confirmed to be 44.5 m/s after addressing algebraic errors.

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  • Knowledge of vector decomposition for projectile motion
  • Familiarity with trigonometric functions, specifically cosine
  • Basic algebra skills for solving equations
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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.
 

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