An object is launched straight upward from ground level with an initial velocity of 32 feet per second. The formula h=32t - 16t ^ 2 gives its height, h, in feet after t seconds. At what time does the object hit the ground?
2 answers:
Answer: B. t=2
Step-by-step explanation:
If we implement t=2 into the function h=32(2)-16(2)^2 we get the height of 0 (Ground level).
Solve for h by simplifying both sides of the equation, then isolating the variable.
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I think it’s D hope that helped
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