a) 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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