Question

Consider firing a pumpkin from a canon at a target that is 17 meters away. The pumpkin follows a quadratic path as displayed to the right. Will the pumpkin hit the target?
I need someone to help me because I'm lost. If you can help, PM me or comment. If not, please don't 

I specifically need to know if B is right and how to do C

Answer 1

A.
(0,00
(2.6,7.9)
(4.8,12.4)
(9.7,15.1)

b.
well, the points don't look like they are on the line but they actually are (plot twist)


so
since (0,0) is on th egarph

0=a(0)²+b(0)+c
0=c
so
f(x)=ax²+bx+0 or
f(x)=ax²+bx

find a and b
sub points

(2.6,7.9)
7.9=a(2.6)²+b(2.6)
7.9=6.76a+2.6b

(4.8,12.4)
12.4=a(4.8)²+b(4.8)
12.4=23.04a+4.8b

use those 2 equations
7.9=6.76a+2.6b
12.4=23.04a+4.8b

eliminate b
multiply first equation by -4.8 and 2nd by 2.6 and add them
-37.92=-32.448a-12.48b
32.24=59.904a-12.48b +
-5.68=27.456a+0b

-5.68=27.456a
divide both sides by 27.456
(-5.68/27.456)=a
find b

12.4=23.04a+4.8b
12.4=(-130.8672/27.456)+4.8b
12.4+(130.8672/27.456)=4.8b
(12.4+(130.8672/27.456))/4.8=b

da equation is
$y= \frac{-5.68}{27.456} x^2+ \frac{12.4+\frac{130.8672}{27.456}}{4.8}x$



c. the roots are found with your calculator
the roots are at x=0 and x=17.287323943662
so very close to the target but not exactly on it
if the target has a radius of 0.287323943662 or more then it will hit the target