anonymous
  • anonymous
A ball is thrown upward with a speed of 40 feet per second from the edge of a cliff 500 feet above the ground. What is the speed of the ball when it hits the ground? Use acceleration due to gravity as –32 feet per second squared and approximate your answer to 3 decimal places.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@zepdrix @IrishBoy123 @SithsAndGiggles
anonymous
  • anonymous
a(t)=-32 v(t)=-32t+C v(0)=-32(0)+C 40=C v(t)=-32t+40 What now?
zepdrix
  • zepdrix
Anti-Differentiate again to get your position function I suppose :)

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More answers

terenzreignz
  • terenzreignz
Did you miss me? :D Integrate it again. To get the position function, y(t).
anonymous
  • anonymous
Of course I did :p Are we not looking for speed? Why would I need position, I did that at first and got an insanely high number.
zepdrix
  • zepdrix
And note that they gave you another piece of initial data for the new unknown constant that shows up: s(0)=500
zepdrix
  • zepdrix
You need to know at what time the object hits the ground, then you can use that time to calculate velocity
anonymous
  • anonymous
I now have -16t^2+40t+00
terenzreignz
  • terenzreignz
@zepdrix so fast T.T
IrishBoy123
  • IrishBoy123
quickest, use \(v^2 = u^2 + 2ax\)
zepdrix
  • zepdrix
So you're looking for the time t when your object is at a height of 0. That is when it hits the ground. 0=-16t^2+40t+500
anonymous
  • anonymous
I meant 500 not 00 btw
zepdrix
  • zepdrix
These guys and their fancy physics shortcuts :) lol
anonymous
  • anonymous
Eh I prefer calc over physics
anonymous
  • anonymous
Leme finish it up.
anonymous
  • anonymous
\[v^2-u^2=2gh,0^2-(40)^2=2*-32*h,h=\frac{ -40*40 }{ -2*32 }=25\] so it goes 25 feet higher. Now total distance =500+25=525 ft initial speed u=o ft/s using above formula again you can find v ,the speed when it touches the ground.
anonymous
  • anonymous
It is position. 500ft cliff above the ground=+500 Thrown upward with speed of 40 feet per second-it will eventually fall down to the ground which is negative=40t acceleration is negative because velocity is negative -32 feet per second squared t is seconds so second squared=-32t^2
anonymous
  • anonymous
Position s=0 means it is a at ground. You find t to know the time it took to reach the ground. You then used the velocity function by taking derivative of position. Plug that time it reached the ground to the velocity function.
anonymous
  • anonymous
∫v(t)=s(t)=−16t2+40t+500
anonymous
  • anonymous
when it comes down g is positive or g=32 ft/s^2
anonymous
  • anonymous
I got t=6.978
anonymous
  • anonymous
Since other solution is negative
IrishBoy123
  • IrishBoy123
well: \(\ddot x = a, \implies \dot x \dfrac{d \dot x}{dx} = a \implies \left|\dfrac{\dot x^2}{2} \right |_u^v = ax \) to meet all those calculus cravings!!
anonymous
  • anonymous
It is canceled correct?
anonymous
  • anonymous
Yall spewing out a bit too much enough
zepdrix
  • zepdrix
t=7? good
anonymous
  • anonymous
Is that the final answer or would I simply take t at 0 and plug it into v(t)?
anonymous
  • anonymous
it should be more than 40|dw:1449793084983:dw|
anonymous
  • anonymous
Which was 7
zepdrix
  • zepdrix
correct, plug it in, v(7)=?
anonymous
  • anonymous
I am getting -184
anonymous
  • anonymous
It would be 184 ft/sec correct?
zepdrix
  • zepdrix
184 for `speed` -184 is the velocity though the negative tells us direction, that it's going down
zepdrix
  • zepdrix
I think the original question asked for velocity, ya?
zepdrix
  • zepdrix
Oh it said speed :) my bad
anonymous
  • anonymous
Nope speed
anonymous
  • anonymous
:)
zepdrix
  • zepdrix
Oh, you've got your units correct also, cool!
anonymous
  • anonymous
hehe
anonymous
  • anonymous
Thanks all
anonymous
  • anonymous
\[v^2-0^2=2*32*525,v=\sqrt{64*525}=8*5\sqrt{21}=40\sqrt{21}\]
IrishBoy123
  • IrishBoy123
@surjithayer +1 !!

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