# what is the maximum height...

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• Oct 10th 2011, 06:06 PM
slapmaxwell1
what is the maximum height...
ok so this problem seems pretty straight forward.

a ball is thrown straight up from the ground with an initial velocity of 256 ft/s. the equation h = 256t -16t^2 describes the height the ball can reach in t seconds.

ok so i take the first derivative and find that t = 8 seconds or in 8 seconds the ball reaches the max height.

when i plugged in the 8 seconds i got 1024 as the max height in feet, my book says the answer is wrong? ok so where did i mess up?
• Oct 10th 2011, 06:47 PM
cheme
Re: what is the maximum height...
Seems correct to me. Even starting with the acceleration function and integrating to velocity you obtain the same answer.
• Oct 11th 2011, 07:40 AM
HallsofIvy
Re: what is the maximum height...
Because that is a simple quadratic you can also find the maximum height by completing the square:
\$\displaystyle h= 256t- 16t^2= -16(t^2- 16t+ 64- 64)= -16(t- 8)^2+ 1024\$
Since a square is never negative, \$\displaystyle -16(t- 8)^2\$ is never positive and h is never larger than 1024 which is its value when t= 8.