Question

a rocket is launched from the top of an 80 - foot cliff with an initial velocity of 88 feet per second. the height of the rocket t seconds after launch is given by the equation h = - 16t²+88t + 80. how high will the rocket be after 2 seconds?

Explanation:

Step1: Substitute t value

Substitute $t = 2$ into $h=-16t^{2}+88t + 80$.
$h=-16\times(2)^{2}+88\times2 + 80$

Step2: Calculate exponent

Calculate $(2)^{2}=4$.
$h=-16\times4+88\times2 + 80$

Step3: Perform multiplications

$-16\times4=-64$ and $88\times2 = 176$.
$h=-64+176 + 80$

Step4: Perform additions

$-64+176=112$, then $112 + 80=192$.
$h = 192$

Answer:

192