anonymous
  • anonymous
A stone is thrown vertically upward with a speed of 27.0 m/s .How much time is required to reach this height? :( plss is there anyone know how to answer it :(
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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GabeBae
  • GabeBae
i'm not sure but i'm pretty sure you would need to use the vertical motion formula
anonymous
  • anonymous
@wowowawa u didn't mention the height, which height?
GabeBae
  • GabeBae
true.. "A stone is thrown vertically upward with a speed of 27.0 m/s .How much time is required to reach this height?" Which height? Are you missing something?

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anonymous
  • anonymous
Specify height.... use formula s=ut+1/2at^
anonymous
  • anonymous
its 26 m
anonymous
  • anonymous
I think you could also use the formula \[\huge v_{fy}^2=v_{0y}^2-2g \Delta y\]Solve for y. You know that the initial velocity is given in the problem, and the final velocity when it reaches top will be 0. This is because in general, the position function is continuous and differentiable, and whenever the position function changes from increasing to decreasing, then the function of the derivative (aka the velocity function) changes from positive to negative. However, we are observing the point in time where it reaches max height. According to Rolle's Theorem, a continuous and differential function with real values must have a certain point in between critical two points that cross the x-axis where the slope/first derivative is 0 (in our case, where the velocity is 0). See the attached picture for a visualization. It might seem complicated, but it should help validate some questions you might have when solving the problem. And to be sure, perhaps @IrishBoy123 can verify
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IrishBoy123
  • IrishBoy123
yes solve using @CShrix 's equation and then get time from \(v = u + at\), or solve the quadratic \(x = ut + \frac{1}{2}at^2\) .... as there will be 2 points at which x = 26, on the way up and on the way back down

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