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# help plz!

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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. Feb 10, 2021

### 1+0 Answers

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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   Feb 10, 2021