Answer :
[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 hₒ 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 vₒ, 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 vₒ=255
then we end up with [tex]\bf h(t)=-16(5)^2+255(5)+0[/tex]
now, we know the cannon is on the ground, thus hₒ 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 vₒ, 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 vₒ=255
then we end up with [tex]\bf h(t)=-16(5)^2+255(5)+0[/tex]
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