H (t ) = -16t^2 + 64t
The maximum height of the ball occurs how many seconds after it was thrown?

Question

H (t ) = -16t^2 + 64t
The maximum height of the ball occurs how many seconds after it was thrown?

anonymous · Answer

-b/2a?

jdoe0001 · Answer

http://openstudy.com/study#/updates/526c1d5fe4b0116e63688fe7

anonymous · Answer

this is different. If I use -b/2a I get 2 seconds because its not asking for the vertex

aakashsudhakar · Answer

Since the derivative of the equation measures the slope of the equation, the maximum height of the ball would be at the time where the slope equals zero and a relative maximum on the function. So you want to find first where H'(t) = 0.

H(t) = -16t^2 + 64t
H'(t) = -32t + 64
64 - 32t = 0
32t = 64
t = 2

So now we know that at the time of two seconds (t = 2), the slope of the ball is zero, which means that the 

H(2) = -16(2)^2 + 64(2)
H(2) = -16(4) + 128
H(2) = -64 + 128
H(2) = 64

The maximum height of the ball is 64 feet, and it occurs at t = 2 seconds.

anonymous · Answer

isn't 16t^2 =256?

aakashsudhakar · Answer

Depends on the value you plug into t. 16t^2 will equal 256 only at t = 4 seconds. (It also works for t = -4 seconds, but it's impossible to go back in time, so only +4 works in this case.)

anonymous · Answer

oh okay that's cool. thank you