idku
  • idku
I need help with this (math-based-physics)
Mathematics
chestercat
  • chestercat
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idku
  • idku
A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. ------------------------------------------------------------ a) What is the speed of the cannonball 3.7 seconds after it was shot? b) How high above the ground is the cannonball 3.7 seconds after it is shot?
idku
  • idku
I found so far: [1] initial speed, \(\rm u=41.4(m/s)\) [2] angle of cannonball with respect to ground, \(\rm \theta=37.15º\) [3] maximum height the cannonball goes above the ground \(\rm x=31.8(m)\) [4] horizontal displacement is \(\rm 167.6(m)\) [5] the entire time of the ball motion \(\rm 5.08(s)\) ------------------------------------------------- I need: [6] speed of the cannonball at, \(\rm t=3.7(s)\) [7] height (vertical displacement) at, \(\rm t=3.7(s)\)
idku
  • idku
@jim_thompson5910 can you help me please when you are free?

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jim_thompson5910
  • jim_thompson5910
I agree with your answers to [1] and [2]. I got the same results. I'm now checking [3]
idku
  • idku
Oh, no need to check. I am entering answers in my pracitce tool. and I got how to do 1-5 all.
idku
  • idku
I have managed myself through 1-5 with a great pain, nd now trying to understand why V=u+at doesn;t work for me. So I am using this: V=u+at {1} t is time {2} a is acceleration due to gravity. {3} where u is the initial velocity -> (vertic. veloc., because for horiz. a=0) V=25(m/s) + (-9.81 m/s\(^2\))(3.7s)=-11.297m/s (and that is wrong)
jim_thompson5910
  • jim_thompson5910
I'm not the best with physics, but I'm guessing you can find the position function and then take the derivative to find the velocity function. After that, plug in t = 3.7 to find the speed at time 3.7 seconds.
idku
  • idku
so if we have gone from 25m/s to 0 m/s how many second would that have took if every second we lose 9.81m/s that is 25/9.81 seconds= 2.55 so that was 1/2 from start to vertex. Now from vertex to end, I get 5.1 better answer for time
jim_thompson5910
  • jim_thompson5910
also, this page might help http://formulas.tutorvista.com/physics/projectile-motion-formula.html it's helped me get refreshed on the projectile formulas
idku
  • idku
Oh, I am also kind of like that. I am good at math, but my physics skills are like a fahrenheit temperature in artarctia during winter season (July-August)
idku
  • idku
Oh, that looks like it can help. tnx will see.
idku
  • idku
I was chatting with a tutor and they asked bucks....
jim_thompson5910
  • jim_thompson5910
oh don't chat there lol that's a bot or a sales rep
jim_thompson5910
  • jim_thompson5910
just close that portion out and focus on the formulas on the page
idku
  • idku
From previous posts I had a formula \(V=u+at\) it is correct. (Velocity is integral of acceleration w/ respect to time, + initial value of our initial velocity u) So, velocity at at t=3.6 should be given by V=25m/s -9.81(m/s\(^2\))•[5.1(s)-3.7(s)]=11.266m/s but that is incorrect.
jim_thompson5910
  • jim_thompson5910
you're only considering the vertical component though
idku
  • idku
so instead of 25m/s I should take 41m/s /
idku
  • idku
?
jim_thompson5910
  • jim_thompson5910
well I'm thinking it would be the result of [1]
jim_thompson5910
  • jim_thompson5910
since that's the composition of the vertical and horizontal speeds
jim_thompson5910
  • jim_thompson5910
I'm not sure though
idku
  • idku
Maybe I can tag @dan815

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