Question

a rocket is launched upward with a velocity of 64 feet per second from the top of a 40 - foot stage. what is the maximum height attained by the rocket? height function answer how to enter your answer (opens in new window) keypad keyboard shortcuts feet

Explanation:

Step1: Identify the height - time formula

The general formula for the height $h(t)$ of an object in vertical - motion under the influence of gravity is $h(t)=-16t^{2}+v_{0}t + h_{0}$, where $v_{0}$ is the initial velocity and $h_{0}$ is the initial height. Given $v_{0} = 64$ feet per second and $h_{0}=40$ feet, so $h(t)=-16t^{2}+64t + 40$.

Step2: Find the time at which the rocket reaches its maximum height

The function $h(t)$ is a quadratic function of the form $y = ax^{2}+bx + c$ with $a=-16$, $b = 64$, and $c = 40$. The time $t$ at which the vertex of the parabola (maximum of the function since $a<0$) occurs is given by $t=-\frac{b}{2a}$. Substitute $a=-16$ and $b = 64$ into the formula: $t=-\frac{64}{2\times(-16)}=\frac{-64}{-32}=2$ seconds.

Step3: Calculate the maximum height

Substitute $t = 2$ into the height function $h(t)=-16t^{2}+64t + 40$. Then $h(2)=-16\times2^{2}+64\times2 + 40=-16\times4 + 128+40=-64 + 128+40=104$ feet.

Answer:

104