Aball is thrown from an initial height of
7
feet with an initial upward velocity of
27/fts
. the ball's height
h
(in feet) after
t
seconds is given by the following.
=h+7−27t16t2
find all values of
t
for which the ball's height is
17
feet.

IF your equation is this: h(t)= -16t^2 + 27t + 7, then you set it equal to 17 and solve for t.  What this does is tells you the times (t) when the ball is exactly 17 feet off the ground.  Because this is a parabola, the ball will go up past the 17 foot mark, then pass it again on the way back down.  This happens at .55 seconds (going up), and 1.13 seconds (coming down).

The answer is 8 apex