Posted by sydney on .
A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 160t.
How many seconds after launch will the rocket be 350 ft above the ground? Round to the nearest hundredth of a second.

math 
Steve,
16t^2 + 160t = 350
just use the quadratic formula to solve for t.