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anonymous

  • one year ago

A stone is thrown vertically upward with a speed of 24.0m/s (a) How fast is it moving when it reaches a height of 13.0m? b) How much time is required to reach this height? c) Why are there two answers to b?

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  1. anonymous
    • one year ago
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    I know there must be an equation to figure out answers a and b, but I don't see one in my notes. I know that the answer to c is because it goes up and goes back down so it passes 13.0m twice.

  2. mathmate
    • one year ago
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    There are two ways to solve this problem, by the law of conservation of energy or using kinematics. (A) By conservation of energy: KE:Kinetic energy : (1/2)mv^2 PE:potential energy : mgh Assuming PE=0 at ground level, then total energy at ground level=mg(0)+(1/2)m(24)^2 total energy at 13m, = mg(13)+(1/2)m(v^2) equate the two energies and solve for v. (B) use kinemaic equation v1=v0+at where v0=initial velocity=24 m/s v1=velocity calculated from (A). a=acceleration due to gravity = -9.81 m/s^2 solve for t.

  3. anonymous
    • one year ago
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    But how do I figure out the velocity from a

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