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simple physics/kinematics question help...
    #7702860 -

ok:
A body is fired vertically upwards with an initial velocity of 100m/s.
(a) how far will it rise
(b) how long will it take to reach maximum height?
(c) what is the time taken before it strikes the ground again.
(d) what is the velocity at impact
---------------------------------------------
(a) 509.7m
(b) 10.2s

i'm not sure what equation to use for (c):
for part (c):
u=0
a=9.81
s=509.7
v=? 0 mybe?

could anyone point me in the right direction on what eqaution to use out of these:
s= ut
v= u+at
s=ut + 0.5*at^2
v^2=u^2+2as


thanks

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Re: simple physics/kinematics question help... [Re: beneath]
    #7702980 -

aww man i used to be so good at these things

hold on let me go get my notes

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Re: simple physics/kinematics question help... [Re: beneath]
    #7703008 -

assuming no air friction

The way down is the reverse from the way up, so d) is 100 m/s

and c) is equal to b) if you count from the highest point, or 2*b) if you want the total time in the air.


Edited by koppie (12/01/07 04:45 PM)

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Re: simple physics/kinematics question help... [Re: koppie]
    #7703109 -

i've tried this:
s= ut + 0.5*a*t^2
s= 0.5*a*t^2
t= sqaure root of (509.7/4.905)

and i get the answer 10.2

bit in the answer section of my book it sais 20.4, which is double, why?

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Re: simple physics/kinematics question help... [Re: beneath]
    #7703129 -

I think they count from the moment of launch.

So 10.2 seconds to go up, and another 10.2 seconds to come down again.


--------------------

"You should view the world as a conspiracy run by a very closely-knit group of nearly omnipotent people, and you should think of that group as yourself and your friends."

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