# Math

A ball is kicked into the air and follows a path described by h(t)=-4.9t^2 +8.4t+0.6, where t is the time in seconds and h is the height of the balls in metres. Determine the maximum height of the ball, to the nearest tenth of a metre.

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1. This is a parabola, and the peak will be halfway between the roots.
so the roots are...
0=4.9t^2 -8.4t-0.6
0 =t^2 -1.71t-0.122
t=(1.71/2 +- sqrt....

so the midpoint, or the max height occures at t=1.71/2=.86 seconds.
then height is h(t)=-4.9t^2 +8.4t+0.6 with t=above.

If you are in calculus, it is much easier, find when it stop rising upward.
dh/dt=0=-9.8t+8.4, solve for t..then solve for h(that t).

1.71+-sqrt(2.92-.488)/2
t= -.86 +-.78
halfway between these roots are -.86, so the beak occurs

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