nasa launches a rocket at t = 0 seconds. its height, in meters above sea - level, as a function of time is…

nasa launches a rocket at t = 0 seconds. its height, in meters above sea - level, as a function of time is given by h(t)=-4.9t^{2}+88t + 423. assuming that the rocket will splash down into the ocean, at what time does splashdown occur? the rocket splashes down after 21.90 seconds. how high above sea - level does the rocket get at its peak? the rocket peaks at meters above sea - level. question help: video 1 video 2

nasa launches a rocket at t = 0 seconds. its height, in meters above sea - level, as a function of time is given by h(t)=-4.9t^{2}+88t + 423. assuming that the rocket will splash down into the ocean, at what time does splashdown occur? the rocket splashes down after 21.90 seconds. how high above sea - level does the rocket get at its peak? the rocket peaks at meters above sea - level. question help: video 1 video 2

Answer

Explanation:

Step1: Identify the function type

The height - function $h(t)=-4.9t^{2}+88t + 423$ is a quadratic function of the form $y = ax^{2}+bx + c$ where $a=-4.9$, $b = 88$, and $c = 423$. The vertex of a quadratic function $y=ax^{2}+bx + c$ has its $t$ - value (for our $h(t)$) given by $t=-\frac{b}{2a}$.

Step2: Calculate the time at which the rocket reaches its peak

Substitute $a=-4.9$ and $b = 88$ into the formula $t=-\frac{b}{2a}$. $t=-\frac{88}{2\times(-4.9)}=\frac{88}{9.8}\approx8.98$ seconds.

Step3: Calculate the height at the peak

Substitute $t=\frac{88}{9.8}$ into the function $h(t)=-4.9t^{2}+88t + 423$. [ \begin{align*} h\left(\frac{88}{9.8}\right)&=-4.9\times\left(\frac{88}{9.8}\right)^{2}+88\times\frac{88}{9.8}+423\ &=-4.9\times\frac{7744}{96.04}+\frac{7744}{9.8}+423\ &=-\frac{37945.6}{96.04}+\frac{7744}{9.8}+423\ &=-395.1+790.2+423\ &=818.1 \end{align*} ]

Answer:

$818.1$