Question 1182626
 NASA launches a rocket at t=0 seconds. It’s height, in meters above sea level, as function of time is given by h(t)=-4.9t^2+208t+339.

solve by formula method the quadratic equation

Roots: 44.02, -1.57
Ignore negative
It will splash after 40.02 s

How high above sea-level does the rocket get at its peak?
Maximum height  occurs when t = -b/(2a) 
 
= -208/((2)(-4.9))
=21.22

f(21.22)  = -4.9*(21.22)^2 +(208 *21.22) + 339 =2546 m

Maximum height = 2546 m above sea level