Question

The height, h, in meters above the ground, of a projectile at any time, t, in seconds, after the launch is defined by the function h(t) = −6t2 + 15t + 2. The graph is shown below. When rounded to the nearest tenth, what is the maximum height reached by the projectile, how long did it take to reach its maximum height, and from what height was the projectile initially launched?

Answer 1

The equation $h(t) =-6t^2 + 15t + 2$ sets the parabola. You can find the parabola vertex as:

$t_v= \frac{-b}{2a} =- \frac{15}{2\cdot (-6)} = \frac{15}{12}= \frac{5}{4}=1.25$

then 

$h( \frac{5}{4} )=-6\cdot (\frac{5}{4} )^2+15\cdot \frac{5}{4} +2= \frac{91}{8} =11.375$

This means that the maximum height is 11.375 m and a projectilereaches this height at 1.25 seconds.

The projectile initially launched at time t=0, so the initial height is h(0)=2 m