Please help....
A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = 0.04x+ 8.3x + 4.3, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
208.02 m 416.03 m 0.52 m 208.19 m
the answer is D right
is that supposed to be "0.04x^2"?
@apoorvk yes...
what did you do here? equated 'y' to 0?
hold on...
. . . -8.3 +- √((8.3^2) - (4)(0.04)(4.3)) x = ----------------------------------------… . . . . . . . . . (2)(0.04) . . . -8.3 +- √68.202 x = ------------------------ . . . . . . . . . 0.08 This results in the following two possible answers. . . . -8.3 + √68.202 x = ---------------------- = -0.52 . . . . . . 0.08 . . . -8.3 - √68.202 x = ---------------------- = -207.36 . . . . . . 0.08
but i wasnt sure of the final result
i THINK its D
well, good work, but how can this -8.3 + sqrt68.... be so large? ever thought about this? sqrt64 is 8, so sqrt68... would be a bit more than 8.
@apoorvk wait so D is the correct answer right?
never mind......
Wait a sec, there seems to be a calculation error.
o...
what was the error @apoorvk
...
ok
sqrt(68.202) = 8.25 something. so, -8.3+8.25.. = something so x = that something^ divided by 0.08 which is again negative. I seem to getting a negative value for 'x'. I wonder if that makes any sense... How can distance be negative?
o...so i did it wrong
No, the calculations that you've shown are correct. But the final thingy seems wrong to me. anyways, even if we consider negative answers, then i seem to be getting 206.95... but options A and D are so proximate then...
thats what i came down to...now i just dont now if its a or d
sorry i have u stuck on this question
Lol, no need to say that. I guess just move on to next problem for now, and clarify this with your teacher later.
yea
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