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idku
 one year ago
I need help with this
(mathbasedphysics)
idku
 one year ago
I need help with this (mathbasedphysics)

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idku
 one year ago
Best ResponseYou've already chosen the best response.0A 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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 can you help me please when you are free?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I agree with your answers to [1] and [2]. I got the same results. I'm now checking [3]

idku
 one year ago
Best ResponseYou've already chosen the best response.0Oh, no need to check. I am entering answers in my pracitce tool. and I got how to do 15 all.

idku
 one year ago
Best ResponseYou've already chosen the best response.0I have managed myself through 15 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
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0also, this page might help http://formulas.tutorvista.com/physics/projectilemotionformula.html it's helped me get refreshed on the projectile formulas

idku
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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 (JulyAugust)

idku
 one year ago
Best ResponseYou've already chosen the best response.0Oh, that looks like it can help. tnx will see.

idku
 one year ago
Best ResponseYou've already chosen the best response.0I was chatting with a tutor and they asked bucks....

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0oh don't chat there lol that's a bot or a sales rep

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0just close that portion out and focus on the formulas on the page

idku
 one year ago
Best ResponseYou've already chosen the best response.0From 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
 one year ago
Best ResponseYou've already chosen the best response.0you're only considering the vertical component though

idku
 one year ago
Best ResponseYou've already chosen the best response.0so instead of 25m/s I should take 41m/s /

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0well I'm thinking it would be the result of [1]

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0since that's the composition of the vertical and horizontal speeds

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure though
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