Asked by sabrina
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.
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.
Answers
Answered by
bobpursley
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
16t^2-250t+730=0
Using the quadratic equation...
t= (250+-sqrt (250^2-4*16*730))/32
solve for t
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