Posted by **danielle** on Sunday, March 25, 2012 at 8:51pm.

A ball is thrown vertically upward with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t)=96t-16t^2

a.) at what time will the ball strike the ground

b.) for what time t is the ball more than 128 feet above the ground?

c.) when will the ball reach its highest peak? how high is it above the ground?

- Math- -
**Damon**, Sunday, March 25, 2012 at 9:32pm
a)

when is s = 0 ?

0 = t(96 - 16 t)

t = 0 of course, it starts at 0

t = 96/16 = 6 seconds

b)

when is s = 128 (two times, on the way up and on the way down)

128 = 96 t - 16 t^2

8 = 6 t - t^2

t^2 - 6 t + 8 = 0

(t-4)(t-2) = 0

so between 2 seconds and 4 seconds

c)where is the vertex of this parabola?

Well it is halfway in time between 2 seconds and four seconds, which is 3 seconds

it is also halfway in time between 0 seconds and 6 seconds, which of course is also at 3 seconds

so

s(3) = 96(3) - 16(9)

= 144 feet

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