edgelordross
contestada

A rock is launched from a cannon. It's height, h (x), can be represented by a quadratic function in terms of time, x, in seconds.

After 1 second, the rock is 243 feet in the air
After 2 seconds, the rock is 452 feet in the air

Find the height, in feet, of the rock after 5 seconds.

Respuesta :

[tex]\bf \textit{initial velocity}\\\\h(t) = -16t^2+v_ot+h_o \qquad \text{in feet}\\\\ v_o=\textit{initial velocity of the object}\\ h_o=\textit{initial height of the object}\\ h=\textit{height of the object at "t" seconds}[/tex]

now, we know the cannon is on the ground, thus hâ‚’ is 0
we also know the -16, in case you're wondering, that's the gravity acceleration in feet divided by 2

what we dunno is vâ‚’, or the initial velocity

however, let's use those values given

[tex]\bf \begin{cases} t=1\\ h(t)=243 \end{cases}\implies 243=-16(1)^2+v_o(1)+0 \\\\\\ 243=-16+v_o\implies \boxed{259=v_o}\\\\ -----------------------------\\\\ \begin{cases} t=2\\ h(t)=452 \end{cases}\implies 452=-16(2)^2+v_o(2)+0 \\\\\\ 452=-64+2v_o\implies \cfrac{516}{2}=v_o\implies \boxed{258=v_o}[/tex]

so, with every passing second, the initial velocity is dropping by 1 foot
after 1 it was 259, then after 2 seconds it became 258

if that continues, after 5 seconds, it'll be 255

so, now if we make t=5 and vâ‚’=255

then we end up with   [tex]\bf h(t)=-16(5)^2+255(5)+0[/tex]
Blacklash

Answer:

The height, in feet, of the rock after 5 seconds = 875 feet

Step-by-step explanation:

Let the quadratic equation h(x) be ax²+bx+c.

             h(x) =  ax²+bx+c

We have

At 0 seconds,the rock is 0 feet in the air.

             h(0) = 0 =  a x 0²+b x 0+c

             c = 0

             h(x) =  ax²+bx

After 1 second, the rock is 243 feet in the air

             h(1) = 243 =  a x 1²+b x 1

             a + b = 243 ------------------------------eqn1

After 2 seconds, the rock is 452 feet in the air

             h(2) = 452 =  a x 2²+b x 2

             4a + 2b = 452 ------------------------------eqn2

eqn1 x 2

             2a + 2b = 486 ------------------------------eqn3

eqn 3 -   eqn2

            2a + 2b - 4a - 2b = 486 - 452

            -2a = 34

               a = -17

Substituting in eqn 1  

             -17 + b = 243

                    b = 260  

So                      h(x) =  -17x²+260x

The height, in feet, of the rock after 5 seconds = h(5) =  -17 x 5²+260 x 5 = 875 feet

The height, in feet, of the rock after 5 seconds = 875 feet

Otras preguntas