Question

The height of a soccer ball is modelled by h(t) = -4.9t^2+ 19.6t + 0.5, where height h(t) is in metres and time, t, in seconds.
a) What is the maximum height of the ball?

Answer 1

h(t) = -4.9t^2 + 19.6t + 0.5
= -4.9(t^2-4t)+0.5

-4 divided by 2 = 2^2 = 4

=-4.9(t^2-4t+4-4)+0.5
=-4.9(t^2-4t+4-4)+0.5
= -4.9(t-4t+4)+0.5+19.6
=-4.9(t-2)^2+20.1

Answer 2

The maximum height is 20.1 m

Answer 3

Is my answer correct?

Answer 4

yup correct

Answer 5

Can I get the same answer with the formula that you posted before?

Answer 6

I think it was....
-2 / ab

Answer 7

another way you could've done was
\(\tt{\dfrac{-b}{2a}}\)

\(\dfrac{-19.6}{2*(-4.9)}\\ \dfrac{-19.6}{-9.8} = 2 \)

\(h(2) = -4.9*2^2+ 19.6*2 + 0.5\\ -4.9*4+39.2\\-19.6 + 39.2 + 0.5 = 20.1\)

Answer 8

yes both ways work
and its

\(\tt{\dfrac{-b}{2a}}\)

Answer 9

Thanks for the help!

Answer 10

anytime \(\ddot\smile\)