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March 31, 2015

March 31, 2015

Posted by **Becky** on Thursday, January 31, 2013 at 9:02pm.

(a) Find the time it hits the ground and (b) the time it reaches its highest point. (c) What is the maximum height?

Please round to two decimal places.

- Pre-Calculus -
**Reiny**, Thursday, January 31, 2013 at 9:15pma) when it hits the ground, f(t) = 0

0 = -16t^2 + 67t

16t^2 - 67t = 0

t(16t - 67) = 0

t = 0 or t = 67/16 = appr 4.19 seconds

t=0 would be the start of the toss

so the answer you want is t = 4.19 sec

b) since you labeled it "pre-calculus" I assume you cannot at this point take the derivative, too bad

so we have to complete the square

f(t) = -16t^2 + 67t

= -16(t^2 - (67/16)t**+ 4489/1024 - 4489/1024**)

= -16( (t - 67/32)^2 - 4489/1024)

= -16(t - 67/32)^2 + 4489/64

it will reach the maximum height at 67/32 or appr 2.09 sec, and that max height is 4489/64 or 70.14 ft

Notice we could have used the properties of the parabola to shorten this up a bit

since the x-intercepts are 0 and 4.19, the vertex must lie midway between them

This would be 2.09 , the answer I got when completing the square. Subbing 2.09 in for f(2.09) gives us

-16(2.09)^2 + 67(2.09) = 70.14

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