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A ball is thrown upward from the roof of a building 60m tall. The ball reaches a height of 80m above the ground after 2s and hits the ground 6s after being thrown.

a)sketch a graph that shows the height of the ball as a function of time (done) (0,60)to(6,0) 

b)state the domain and range D={xer|0<=x<=6} R={yer|0<=y<=80}

This one I dont get

c) Determine an equation for the function.

I used f(x)=a(x-2)2+80 but when I input a point the answer is never correct. 

 
Baneofwizard  Oct 10, 2018
 #1
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You need to know the basic equation of ballistic motion

 

\(h(t) = -\dfrac{g t^2}{2} + v_0 t + h_0\)

 

here they give you \(h_0 = 60\)

 

they also seem to be using \(g = -10 ~m/s^2\) which isn't quite accurate but close enough for government work.

 

You have two equations

 

\(h(2)= 80,~h(6) = 0\\ \text{you really only need one of them, plugging in values we have}\\ 80 = -5(2^2) + 2v_0 + 60\\ 20 = 2v_0 -20\\ v_0 = 20 \\ \text{checking with the other equation we have}\\ h(6) = -5(6^2) + 20(6) + 60 = \\ -180 + 120 + 60 = 0 \text{ as expected. So our equation of motion is}\\ h(t) = -5t^2 +20t + 60 \\ \text{we can rewrite this as}\\ h(t) = -5(t^2 - 4t - 12) = \\ -5((t-2)^2-4-12)=\\ -5(t-2)^2 + 80 \\ \text{and this last is the standard form of the parabola }\)

 
Rom  Oct 10, 2018

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