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

You are trying to find the maximum of the function, so take the first derivative of the function h(t), solve for t where $$ \frac{d}{dt} h(t) = 0 $$, then substitute t into the function h(t) to get the answer.

Alternatively you can graph the function h(t) out and find the maximum. This is easily done with a graphing calculator.

anonymous · Answer

I still can't figuer it out, can you show me how please to set up the equation.

anonymous · Answer

$$ h(t) = -4.9t^2 +19.6t + 58.8$$
$$ \frac{d}{dt} h(t) = h'(t) = -9.8t+19.6 $$
$$ h'(t) = 0 \rightarrow 9.8t = 19.6 $$
$$ t = \frac{19.6}{9.8} $$
$$ h_{max} = h(\frac{19.6}{9.8})$$

anonymous · Answer

Can you tell me if I am correct in this please?
hmax = 58.8(2) = 117.6 meters   is max height

anonymous · Answer

answer is 78.4 meters