<PRE><B>Question 613244</B>
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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Sub 0.8 for t
h(0.8) = -16*0.8^2 + 10*0.8 + 40
= 37.76 feet
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When will the ball be 20 feet above the ground and headed down?
h = 20
20 = -16t^2 + 10t + 40
{{{-16t^2 + 10t + 20 = 0}}}
*[invoke solve_quadratic_equation -16,10,20]
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Ignore the negative solution.



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