An object is launched at 19.6 m/s from a height of 58.8 m. the equation for the height (h) in terms of time (t) is given by h(t) = -4.9t2 +19.6t + 58.8. what is the object's maximum height?

H(t) = Ho +Vot - gt^2/2

Vo = 19.6 m/s
Ho = 58.8 m
g = 9.8 m/s^2

H(t) = 58.8 + 19.6t -9.8t^2/2 = 58.8 + 19.6t - 4.9t^2

Maximun height is at the vertex of the parabole

To find the vertex, first find the roots.

58.8 + 19.6t - 4.9t^2 = 0

Divide by 4.9

12 + 4t - t^2 = 0

Change sign and reorder

t^2 - 4t -12 = 0 

Factor

(t - 6)(t + 2) =0 ==> t = 6 and t = -2.

The vertex is in the mid point between both roots

Find H(t) for: t = [6 - 2]/2 =4/2 = 2

 Find H(t) for t = 2

H(6) = 58.8 + 19.6(2) - 4.9(2)^2 = 78.4

the maximum height is 78.4 m

Its d so thats how it would would back from 1.0 form three four two

The answer should be 5.