Holly fired a cannon. The cannon ball was shot from 2 feet above the ground. 0.1 seconds later, it was 33 feet above the ground. At 0.2 seconds, it was 22 feet above the ground.The cannon ball’s trajectory follows a parabolic path: f(x) = ax^2 + bx + c
Determine the coefficients a, b, and c.
+128408 Let x be the number of seconds that have elapsed
Since at x = 0 the cannon ball was 2 ft off the ground, then c = 2
And we have that
a(.1)^2 + b(.1) + 2 = 33 ⇒ .01a + .1b = 31 (1)
And
a(.2)^2 + b(.2) + 2 = 22 ⇒ .04a + .2b = 20 (2)
Multiply (1) by -2 ⇒ -02a - .2b = -62
Add this to (2)
.02a = -42 divide both sides by .02
a= -2100
And using any equation , we can find b as
.04 (-2100) + .2b = 20
-84 + .2b = 20
.2b = 20 + 84
.2b = 104 divide both sides by .2
b = 520
a = -2100
b = 520
c = 2
See the graph, here : https://www.desmos.com/calculator/sjtjtjnlwc
Second part .....I think they mean the Quadratic Formula....not the Pythagorean Theorem
Time to hit the ground =
(-520 ± sqrt ( 520^2 - 4 (-2100) (2) ) ) / ( 2 * -2100) =
(-520 ± sqrt (287200) ) /( -4200) =
Take the negative sqrt
(-520 - 20sqrt (718) ) / (-4200) ≈ .25 seconds
