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h(t) = -9.8t2 – 10t + 330, where t is in seconds, and height is in meters. After...

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shutto52 | Student, College Freshman | eNotes Newbie

Posted July 30, 2011 at 3:46 AM via web

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h(t) = -9.8t2 – 10t + 330, where t is in seconds, and height is in meters. After how many seconds will the stone hit the ground?

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beckden | High School Teacher | (Level 1) Educator

Posted July 30, 2011 at 3:55 AM (Answer #1)

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You want to solve

-9.8t^2 - 10t + 330 = 0

Use the quadradic formula

`t = (-b+-sqrt(b^2-4ac))/(2a)`

`t=(-(10)+-sqrt((10)^2-4(-9.8)(330)))/(2(-9.8))=(-10+-114. 175)/(-19.6)`

t = 6.335 sec  or -5.315 sec.

Obviously time is only positive so the answer is 6.335 seconds

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shaznl1 | High School Teacher | (Level 1) Salutatorian

Posted July 30, 2011 at 10:20 AM (Answer #2)

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well the stone hits the ground when the height = 0

And here you have given a height function

so we only need to set it equal to zero.

Using quadratic formula ((-b+/-sqrt(b^2-4ac))/2a

plugging in all the numbers

a= -9.8 b- -10 c= 330

the answer is 6.335 or -5.315 and the unit is seconds.

Since time really couldn't be negative, (time machine?)

 the only legitmate answer to this question is 6.335 seconds

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