Question

A rock is launched from a cannon. After 1 second, the rock is 243 feet in the air; after 2 seconds, it is 452 feet in the air. Using the quadratic function, find the height, in feet, of the rock after 5 seconds in the air

Answer 1

I've seen a similar problem to it yesterday, and I got confused lol ^_^"

Answer 2

What is the angle at which it is launched?

Answer 3

doesn't say.

Answer 4

a+b = 243
a = 243 - b

4a +2b = 452
4(243-b) +2b = 452

892 -4b + 2b = 452

892 -452 = 2b

440/2 = b

b = 220
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a = 243-220 = 23

23x^2 +220x = y
--------------------

23 + 220 = 243 checks

23(4) + 220(2) = 452
92 + 440 = 30
ok.....whatd I do wrog..hard to type in the dark :)

Answer 5

assume it is launched straight up. say its initial acceleration is a. distance is ut+1/2at^2

Answer 6

initial speed is 0. so d = 1/2 at^2

Answer 7

its a simple quad equation, not physics....but that might be a shot :)

Answer 8

(0,0) (1,243) (2,452)

Answer 9

since it is launched straight up in the air, the gravity is acting against the rock. so s = 1/2(a-32.17)t^2

Answer 10

1a + 1b = 243
b = 243-a

4a + 2b = 452
4a + 2(243-a) = 452

4a +486 -2a = 452

2a = 452 - 486
2a = -34

a = -17 by that

Answer 11

oh okay, then you should follow amistre 64's procedure.

Answer 12

b = 243 --17
b = 260

-17x^2 +260x = y

-17(4) + 260(2) = 452
-68 + 520 = 452 right? someone tell me its right lol

Answer 13

452 is the height of the rock at 2 seconds.

Answer 14

amistre64 can u please help me

Answer 15

the numbers are working yay!!...plug in a 5

Answer 16

plug in a 5 where?

Answer 17

-17(25) +260(5) = ?

Answer 18

that's what i thought. i just tried it. i got 875

Answer 19

-425 + 1300= 875....me too :)

Answer 20

correct! one more to go! let's hope i pass it this time! i've only missed one outta 4. on to the 5th question. thanks amistre64!

Answer 21

i wrote you a quad program today, but left my flash drive at the college lol

Answer 22

thanks though!