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April 1, 2015

April 1, 2015

Posted by **sammi** on Sunday, February 21, 2010 at 5:51pm.

This type of problem always shows up on my test and I always get it wrong. Can someone show me how to do it? Thanks.

- calculus -
**Reiny**, Sunday, February 21, 2010 at 5:59pmAt this point you should know that

if s(t) is the height obtained, then

v(t) , or velocity, is s'(t) and

a(t) , the acceleration, is v'(t) or s''(t)

Also having done some of these, by now you should know that for the above,

s(t) = -16t^2 + 64t + 480 and

v(t) = -32t + 64

so when it hits the ground, isn't s(t) = 0 ??

so

-16t^2 + 64t + 480 = 0

t^2 - 4t - 30 = 0

t = 2 ± √34 , we can reject the negative value

t = 2+√34

sub that into v(t) and you are done

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