posted by Becky on .
A pomegranate is thrown from a ground level straight up into the air at time t = 0 with velocity 67 feet per second. Its height at time t seconds is f(t)=-16^2+67t.
(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.
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