Question

an object is thrown straight upward. It's height in feet (y) after seconds is given by y=-6t2+128t. What is the maximum height reached by the object? How long does it take to reach that height?

Answer 1

easiest way is to use t=-b/(2a) then solve for y
b=128
a=-6
so t=-128/(2*-6)=32/3
no solve for y
y=-6(32/3)^2+128(32/3)
y=(-2048/3)+(4096/3)
y=(4096-2048)/3
y=2048/3