Rocket Height Model and Vertex Interpretation
A model rocket is launched from a 5-foot platform. Its height in feet, t seconds after launch, is modeled by $h(t)=-16t^2+96t+5$ for $0\le t\le 7$. A second rocket uses $h_2(t)=-16t^2+80t+5$. The team wants to compare maximum heights, estimate when the first rocket returns to the ground, and determine whether raising the launch platform by 3 feet changes the time at which the first rocket reaches its maximum height. All times may be rounded to the nearest tenth when appropriate.
1. Find the maximum height of the first rocket and the time at which it occurs.
Read the worked answer
The height is a quadratic with $a=-16$, $b=96$. The vertex time is
$$t=-\frac{b}{2a}=-\frac{96}{2(-16)}=3.$$
Substitute $t=3$:
$$h(3)=-16(9)+96(3)+5=-144+288+5=149.$$
The first rocket reaches a maximum height of 149 feet at 3 seconds.
2. Solve for when the first rocket returns to ground level, rounded to the nearest tenth of a second.
Read the worked answer
Set $h(t)=0$:
$$-16t^2+96t+5=0.$$
Using the quadratic formula, the discriminant is
$$96^2-4(-16)(5)=9216+320=9536.$$
Thus
$$t=\frac{-96\pm\sqrt{9536}}{-32}.$$
Since $\sqrt{9536}\approx 97.65$, the positive solution is
$$t\approx \frac{-96-97.65}{-32}\approx 6.05.$$
Rounded to the nearest tenth, the rocket returns to ground level at about 6.1 seconds.
3. Compare the maximum heights of the two rockets, and explain whether raising the first rocket’s platform from 5 feet to 8 feet changes the time of its maximum height.
Read the worked answer
For the second rocket, the vertex time is
$$t=-\frac{80}{2(-16)}=2.5.$$
Its maximum height is
$$h_2(2.5)=-16(2.5)^2+80(2.5)+5=-100+200+5=105.$$
The first rocket’s maximum is 149 feet, so it goes 44 feet higher than the second rocket.
Changing the platform height changes only the constant term: $h(t)=-16t^2+96t+8$. The vertex time depends on $b$ and $a$, not the constant, so it remains $t=3$ seconds. Raising the platform changes the maximum height but not the time at which it occurs.