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) = -4.9t^2 +19.6t + 58.8. What is the object's maximum height?
A. 58.8
B. 194.04
C. 78.4
D. 117.6

Question

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) = -4.9t^2 +19.6t + 58.8. What is the object's maximum height?
	A.	58.8
	B.	194.04
	C.	78.4
	D.	117.6

anonymous · Answer

The easiest way to find Ymax is to use x=-b/2a. That will give you the x value when Y is max. In this case, it is -19.6/-9.8 = 2. Plug in 2 to your equation to get Ymax.