Question - Determining the Time and Maximum Height of a Projectiles Launch

To determine the time at which the projectile reaches its maximum height, we need to find when the velocity is zero. The velocity is the derivative of the position function \( y(t) \).

The position function given in the question is:

\[ y(t) = -16t^2 + 64t + 80 \]

To find the velocity, we differentiate \( y(t) \) with respect to \( t \):

\[ v(t) = y'(t) = \frac{d}{dt}(-16t^2 + 64t + 80) = -32t + 64 \]

We set the velocity to zero and solve for \( t \):

\[ 0 = -32t + 64 \]

\[ 32t = 64 \]

\[ t = 2 \]

The projectile reaches its maximum height at \( t = 2 \) seconds.

Now we calculate the maximum height by substituting \( t = 2 \) back into the position function \( y(t) \):

\[ y(2) = -16(2)^2 + 64(2) + 80 \]

\[ y(2) = -16(4) + 128 + 80 \]

\[ y(2) = -64 + 128 + 80 \]

\[ y(2) = 144 \]

Therefore, the maximum height reached by the projectile is 144 feet.