posted by jatin .
A model rocket is launched with an initial velocity of 250 ft per second. The height h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after the launch will the rocket be 500 ft above the ground? Round to the nearest hundredth of a seconds?
-16t2 + 250t = 500
solve for t
You have to set this equal to zero.
16t^2 -250t + 500 = 0
Now, this isn't factorable. You have to use the quadratic equation to solve for t to the nearest hundredth.
Quadratic x = opposite of b (plus or minus) the square root of( b squared minus 4ac). All of this divided by 2a.