SOLUTION: 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 s
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h = −16t2 + 250t.
How many s
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Question 952389: 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 870 ft above the ground? Round to the nearest hundredth of a second. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! 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 870 ft above the ground? Round to the nearest hundredth of a second.
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-16t^2+250t-870=0
solve for t using quadratic formula:
a=-16, b=250, c=-870
ans: the rocket be 870 ft above the ground:
after 5.23 sec on the way up
after 10.39 sec on the way down
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