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

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.

Answer 1

730=-16t^2+250t

16t^2-250t+730=0

Using the quadratic equation...

t= (250+-sqrt (250^2-4*16*730))/32

solve for t