(h)t=-4.9t^2+98t+8 Find the max height?

Answer 1

are you taking Calculus ?
then

h'(t) = -9.8t + 98 = 0
t = 10
h(10) = -4.9(100) + 98(10) + 8 = 498

or

complete the square

h(t) = -4.9(t^2 - 20t + 100 - 100) + 8
= -4.9( (t-10)^2 - 100) + 8
= -4.9(T-10)^2 + 490+8
= -4.9(t-10)^2 +498

vertex is (10, 498) , so max height is 498