Question

The function h=14t²+56t+60 represents the height of an object that is launched directly up into the air, where h represents the height in feet of the object and t represents time in seconds
What is the maximum height in feet of the object?
How long will it take the object to reach its maximum height?

Answer 1

Answer:

It will take the object 2 seconds to reach the maximum height.

Step-by-step explanation:

The original equation is $h=-14t^2+56t+60$.

The object reaches the maximum height when derivative of $h$ is zero

$\frac{dh}{dt}=0$

So let us take evaluate the derivative of $h$

$\frac{dh}{dt} =\frac{d}{dt} (-14t^2+56t+60)=\frac{d}{dt} (-14t^2)+\frac{d}{dt}(56t)+\frac{d}{dt} (60)$

$=-28t+56$

this must be equal to 0:

$-28t+56=0$

$t=\frac{-56}{-28}$

$\boxed{ t=2}$

It will take the object 2 seconds to reach the maximum height.