Using the voluma given h=-16t²<span>+40t+3, we can find the seconds after the h=10 by subsituting 10 as h, then isolating t
</span>h=-16t²+40t+3
10=-16t²+40t+3 h=10 was inserted
-16t²+40t+3-10=0 10 was moved to the right side
-16t²+40t-7 Like terms were collected
-(16t²-40t+7) Factored out -1 (completely optional)
Then complete the square
-(16t²-40t)-7 Moved -7
-4(2t²-10t)-7 Factored out 4
-8(t²-5t)-7 Factored out 2
-8(t²-5t+
)-7 +
-8(t-5/2)² -7+
-8(t-5/2)² -7+
-8(t-5/2)²-
OR
You could use the QUADRATIC FORMULA
(Legit so fun....)
t=
(a=16, b=-40, c=7)
t=
t=
t=
1) Time is therefore
and
When finding the maxumium height, there is a rule that when h
is achieved the velocity is ALWAYS ZERO. Therefore when graphing the path of the ball the vertex's y-coordinate will be the maximum height, and the x-coordinate is the value of time at that particular height.
2) The maximum height is 28 meters reached at 1.25 seconds