Question

problem #2 on mars, gravity is less than it is on earth. if you were to kick a mars rock at an initial velocity of 56 feet per second from the top of 30 - foot hill, the rocks height h would be modeled by the equation h(t)=-1.9t² + 56t + 30 where t is time in seconds. a: how high would the rock be after 20 seconds? b: how high would the rock get?

Explanation:

Step1: Solve for part A

Substitute $t = 20$ into the height - function $h(t)=-1.9t^{2}+56t + 30$.
$h(20)=-1.9\times(20)^{2}+56\times20 + 30$

Step2: Calculate the powers and multiplications

First, calculate $(20)^{2}=400$. Then, $-1.9\times400=-760$, and $56\times20 = 1120$.
$h(20)=-760+1120 + 30$

Step3: Calculate the sum

$h(20)=(-760+1120)+30=360 + 30=390$

Step4: Solve for part B

The function $h(t)=-1.9t^{2}+56t + 30$ is a quadratic function of the form $y = ax^{2}+bx + c$ where $a=-1.9$, $b = 56$, and $c = 30$. The $t$ - value of the vertex of a quadratic function is given by $t=-\frac{b}{2a}$.
$t=-\frac{56}{2\times(-1.9)}=\frac{56}{3.8}\approx14.74$

Step5: Find the maximum height

Substitute $t=\frac{56}{3.8}$ into the function $h(t)$.
$h(\frac{56}{3.8})=-1.9\times(\frac{56}{3.8})^{2}+56\times\frac{56}{3.8}+30$
$h(\frac{56}{3.8})=-1.9\times\frac{3136}{14.44}+ \frac{3136}{3.8}+30$
$h(\frac{56}{3.8})=-\frac{5958.4}{14.44}+\frac{3136}{3.8}+30$
$h(\frac{56}{3.8})\approx - 412.77+825.26+30$
$h(\frac{56}{3.8})\approx442.49$

Answer:

A. The height of the rock after 20 seconds is 390 feet.
B. The rock would get approximately 442.49 feet high.