a stone is tossed from a bridge into the river below. its path is represented by h=-2t^2+7t+9 where h is height(in metres) and t is time (in seconds)

a) what is the maximum height and how many seconds does it take to reach the maximum height?

b) how many seconds does it take for the stone to reach the water?

2. Originally Posted by euclid2
a stone is tossed from a bridge into the river below. its path is represented by h=-2t^2+7t+9 where h is height(in metres) and t is time (in seconds)

a) what is the maximum height and how many seconds does it take to reach the maximum height?

b) how many seconds does it take for the stone to reach the water?
to a):

The maximum height is the y-cooordinate of the vertex:

$\displaystyle h(t)=-2(t^2 - \frac72 t \bold{\color{red} + \frac{49}{16}}) \bold{\color{red} -\left(- \frac{49}{8}\right)} + 9 = -2(t-\frac74)^2 +\frac{121}8$

The vertex is at $\displaystyle V(\frac74\ ,\ \frac{121}8)$

to b):

The stone has reached the surface of the water if h(t) = 0. Solve for t:

$\displaystyle -2t^2+7t+9=0$

You'll get 2 solutions but only one of them is plausible with your problem.
(For confirmation only: $\displaystyle t = \frac92$ )