+203 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.
+6251 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
\(h(2)= 80,~h(6) = 0\\ \text{you really only need one of them, plugging in values we have}\\ 80 = -5(2^2) + 2v_0 + 60\\ 20 = 2v_0 -20\\ v_0 = 20 \\ \text{checking with the other equation we have}\\ h(6) = -5(6^2) + 20(6) + 60 = \\ -180 + 120 + 60 = 0 \text{ as expected. So our equation of motion is}\\ h(t) = -5t^2 +20t + 60 \\ \text{we can rewrite this as}\\ h(t) = -5(t^2 - 4t - 12) = \\ -5((t-2)^2-4-12)=\\ -5(t-2)^2 + 80 \\ \text{and this last is the standard form of the parabola }\)
