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Mathematics 9 Online
OpenStudy (chrisplusian):

anyone good at kinematic equations from physics?

OpenStudy (chrisplusian):

a football is kicked and it takes 4.4s to hit the ground. What is the maximum height the ball reaches? assume the ball starts at ground level

OpenStudy (anonymous):

distance = velocity times time

OpenStudy (anonymous):

integrating to get acceleration gives 1/2at^2 where a is acceleration and t is time

OpenStudy (anonymous):

the acceleration due to gravity is 9.81 m/s^2

OpenStudy (anonymous):

so the distance = (1/2)(9.81)(4.4)^2 = 94.960 m

OpenStudy (chrisplusian):

I was never told you could intergrate to get acceleration from that function!!!!??????? I just doulble checked my notes and it doesn't say that. Plus their answer was ymax=24m

OpenStudy (anonymous):

lol wel i have clearly gone wrong. what formula did they use?

OpenStudy (anonymous):

yeah ive done this stupidly

OpenStudy (anonymous):

so you dont know the initial velocity?

OpenStudy (anonymous):

no wait, of course you dont need it

OpenStudy (anonymous):

you want y when t = 2.2 with a = 9.81

OpenStudy (chrisplusian):

no all I know is four kinematic equations and the fact that the acceleration is-9.8 (m/s^2) and that the final velocity would equal zero, aside from time equaling zero

OpenStudy (anonymous):

i feel like ivanmlerner is going to show us the way

OpenStudy (chrisplusian):

@JamesWolf when I used the equations I know I came up with the same answer you did I don't know how or why. But it is wrong. I wonder how did you come up with 2.2=t?

OpenStudy (anonymous):

wel ball reaches maximum height halfway through the time?

OpenStudy (chrisplusian):

is that always true?

OpenStudy (anonymous):

it needs to be at its maximum height at 2.2s, then, since h(t)=(1/2)*gt^2, just put t = 2.2, you were in the right way @JamesWolf

OpenStudy (anonymous):

i think...|dw:1350000898536:dw|

OpenStudy (anonymous):

im not sure if its always true, damn i was so close!

OpenStudy (anonymous):

hahaha im wrong too

OpenStudy (anonymous):

no, that gives 23.74 = 24

OpenStudy (anonymous):

oh wait auhuha its right

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

Of course, because it needs to fall from rest at the max point, during 2.2 sec at aceleration g, no initial velocity and no y0

OpenStudy (anonymous):

I just put the origin at the max point.

OpenStudy (anonymous):

presuming its terminal velocity is infinate

OpenStudy (anonymous):

What do you mean? sorry not native speaker.

OpenStudy (anonymous):

no thats fine. i mean the highest point will only be halfway through its flight, if the velocity up is the same as the velocity down

OpenStudy (anonymous):

if the velocity down is less, because say it gets slowed by air resistance. then the highest point will be earlier in time

OpenStudy (chrisplusian):

I understand the concept But I am asked to show this using the kinematic equations, which I can't. give me a minute and I will show you all the eqations I am supposed to use

OpenStudy (anonymous):

okay

OpenStudy (chrisplusian):

\[x=\Delta t\]

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