How fast would a ball have to be thorwn upward to reach a maximum height of 256ft? [Hint: Use the discriminant of the equation 16t^2-v0t + h=0
Stacey Warren - Expert brainly.com
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"how fast" and "have to be thrown" imply that we seek the absolute minimum velocity to achieve this result. This should suggest to the mind the vertex of a downward opening parabola. This being the case, I have to question your equation. I would expect to see -16, not +16.
I copied this question straight from my book its definitely the correct equation for the problem. I need the maximum velocity to achieve a result.
No, you need the minimum velocity to achieve the result. If you shoot the ball out of a canon at 4500 ft/sec, you will easily achieve 256 ft.
Anyway, I guess we get to assume h = 0? We'll be throwing from the ground?
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oh okay you make more sense now and yes I believe so
16t^2-v0t + h=0
So, our new equation is just
We just need to know when this thing just makes it to 256 feet. Since this is the graph of a parabola, that should be easy enough a couple of ways. Shall we use the geometry or pull out the calculus? Your choice.
Perfect. With 16t^2 - v0(t) = f(t), we have two points (values for t) where this function is zero.
16t^2 - v0(t) = t(16t - v0) = 0, and this t = 0 and t = v0/16. Make sense?
yes it makes sense
Excellent. What we know about parabolas is that the vertex (highest or lowest point), is exactly between these two zeros. You may have seen the expression -b/(2a).
Thus, t = (0 + v0/16)/2 = v0/32. This is the TIME at which the ball reaches the vertex of the parabolic flight. We still don't know how high it goes, we just know WHEN it gets there.
Still absorbing it all?
yes, its all coming together
Only a little more algebra to go.
We have f(t) = 16t^2 - v0t
We need f(v0/32) = 256. Can you solve it?
yes I think so, I got..
Hopefully, you also found -128 ft/sec and discarded it?