FUCK MATH, GODDAMNIT!!!

I hope math drives its car off a fucking cliff!

The car is travelling at 38 km/h at the time it meets the edge of the cliff.  The cliff is 120 m high.

What distance from the base of the cliff will the car land?  Also, please provide the quadratic equation that describes the parabolic path of the car.

>>1

Why not just try modding out by the ideal generated by math?

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>>3
being up and right our +:
S = (38/3.6)*sqrt(g/240) meters is the distance from the base of the cliff.

(g = gravity acceleration)

also, the quadratic equation can be easily aquired by realising the answer above is one of 2 - simetric - roots of it.

so y = (x - (38/3.6)*sqrt(g/240)) * (x + (38/3.6)*sqrt(g/240))

there ya go

yowsa!

>>6
It takes 3 points to define a nontrivial parabola.

y = A(x - (38/3.6)*sqrt(g/240)) * (x + (38/3.6)*sqrt(g/240))
is an infinite set of parabolas with the same roots.  Are you sure the parabola you stated has the same peak location?

>>8

You're right, we want an A that makes the c (as in ax^2+bx+c) become 120

the C will be -((38/3.6)^2 * (g/240))*A

being that we want that to be 120, A = -120/((38/3.6)^2 * (g/240))

so de correct equation is
y = -120/((38/3.6)^2 * (g/240))*(x - (38/3.6)*sqrt(g/240)) * (x + (38/3.6)*sqrt(g/240))

the*

*headpalm*

ha ha ha, oh wow.

This thread hurts a lot.

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