Question 613244
A ball is thrown off the top of a classroom building with an initial upward velocity of 10 feet per second. The height above the ground can be found using h= -16t(squared) + 10t + 40 with t representing time in seconds and h representing height in feet. How high is the ball after 0.8 seconds? 
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given:
{{{h= -16t^2 + 10t + 40}}}
set t to 0.8 and solve for h:
{{{h= -16(0.8)^2 + 10(0.8) + 40}}}
{{{h= -16(0.64) + 8 + 40}}}
{{{h= -10.24 + 8 + 40}}}
{{{h= -2.24 + 40}}}
{{{h= 37.76}}} feet
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When will the ball be 20 feet above the ground and headed down?
set h to 20 and solve for t:
{{{h= -16t^2 + 10t + 40}}}
{{{20= -16t^2 + 10t + 40}}}
{{{0= -16t^2 + 10t + 20}}}
{{{0= 16t^2 - 10t - 20}}}
{{{0= 8t^2 - 5t - 4}}}
Applying the "quadratic formula" we get:
x = {-0.46, 1.09}
answer:
x = 1.09 secs
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details of quadratic follows:
*[invoke quadratic "x", 8, -5, -4 ]