Question 952389
A model rocket is launched with an initial velocity of 250 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by 
h = −16t2 + 250t.
How many seconds after launch will the rocket be 870 ft above the ground? Round to the nearest hundredth of a second.
***
-16t^2+250t-870=0
solve for t using quadratic formula:
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=-16, b=250, c=-870
ans:  the rocket be 870 ft above the ground:
after 5.23 sec on the way up
after 10.39 sec on the way down