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?

Question

campbell_st · Answer

easiest way to solve this problem is the find the line of symmetry of the curve. The maximum height will be on that line

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for the general equation 

$$y = at^2 + bt + c $$

to find the line of symmetry use 

$$t =-\frac{b}{2a}$$

$$t = \frac{-19.6}{2 \times - 4.9}$$

this is the time when the maximum height occurs. 

Substitute it into the original equation to find the maximum height.