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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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?

Mathematics
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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.
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.
But how do I figure out the velocity from a

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