Using the voluma given h=-16t²+40t+3, we can find the seconds after the h=10 by subsituting 10 as h, then isolating t
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+[tex] \frac{25}{4} [/tex])-7 +[tex] \frac{25}{4} [/tex]
-8(t-5/2)² -7+ [tex] \frac{25}{4} [/tex]
-8(t-5/2)² -7+ [tex] 6\frac{1}{4} [/tex]
-8(t-5/2)²-[tex] \frac{3}{4} [/tex]
OR
You could use the QUADRATIC FORMULA
(Legit so fun....)
t=[tex] \frac{-b \frac{+}{}\sqrt{b^2-4ac} }{a^2} [/tex] (a=16, b=-40, c=7)
t=[tex] \frac{-(-40) \frac{+}{}\sqrt{-40^2-4(16)(7)} }{2(16)} [/tex]
t= [tex] \frac{40\frac{+}{}\sqrt{1600-448} }{32} [/tex]
t= [tex] \frac{40\frac{+}{}\sqrt{1152} }{32} [/tex]
1) Time is therefore [tex] \frac{40+\sqrt{1152} }{32} [/tex] and [tex] \frac{40-\sqrt{1152} }{32} [/tex]
When finding the maxumium height, there is a rule that when h[tex]_{max} [/tex] 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
