Question

a toy rocket is launched into the air. the height, in feet, of the rocket is modeled by the function h(t)= - 16t^2 + 64t + 80, where t is the time, in seconds, that the rocket is in the air. the initial height of the rocket is
, the rocket reaches a maximum height of
after

Explanation:

Step1: Find the initial height

The initial height is when \(t = 0\). Substitute \(t=0\) into the function \(h(t)=- 16t^{2}+64t + 80\).

$$h(0)=-16(0)^{2}+64(0)+80=80$$

Step2: Find the time at which maximum height occurs

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate (in our case \(t\)) of the vertex is given by \(t=-\frac{b}{2a}\). Here \(a=-16\) and \(b = 64\).

$$t=-\frac{64}{2\times(-16)}=-\frac{64}{-32}=2$$

Step3: Find the maximum height

Substitute \(t = 2\) into the function \(h(t)=-16t^{2}+64t + 80\).

$$h(2)=-16\times(2)^{2}+64\times2 + 80=-16\times4+128 + 80=-64+128+80=144$$

Answer:

The initial height of the rocket is 80 feet, the rocket reaches a maximum height of 144 feet after 2 seconds.