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

Question

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 730 ft above the ground? Round to the nearest hundredth of a second.

Reiny · Answer

solve 730 = -16t^2 + 250t 16t^2 - 250t + 730 = 0 (use the quadratic equation formula)