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- idku

I need help with this
(math-based-physics)

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- idku

I need help with this
(math-based-physics)

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- 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

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

@jim_thompson5910 can you help me please when you are free?

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- jim_thompson5910

I agree with your answers to [1] and [2]. I got the same results. I'm now checking [3]

- idku

Oh, no need to check. I am entering answers in my pracitce tool. and I got how to do 1-5 all.

- 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

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

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

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

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

Oh, that looks like it can help. tnx will see.

- idku

I was chatting with a tutor and they asked bucks....

- jim_thompson5910

oh don't chat there lol
that's a bot or a sales rep

- jim_thompson5910

just close that portion out and focus on the formulas on the page

- 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

you're only considering the vertical component though

- idku

so instead of 25m/s I should take 41m/s /

- idku

?

- jim_thompson5910

well I'm thinking it would be the result of [1]

- jim_thompson5910

since that's the composition of the vertical and horizontal speeds

- jim_thompson5910

I'm not sure though

- idku

Maybe I can tag @dan815

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