mohamedhani811 mohamedhani811
  • 19-09-2019
  • Physics
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2. A stone is thrown vertically upward with a speed of 22m/s.
a. How fast is it going when it reaches a height of 25m?

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Chenk13
Chenk13 Chenk13
  • 19-09-2019

Answer:

Explanation:

Energy E is conserved:

[tex]E=\frac{1}{2}mv^2+mgh[/tex]

If v₀ = 22m/s, h₀=0m and h₁=25m:

[tex]E=\frac{1}{2}mv_0^2=\frac{1}{2}mv_1^2+mgh_1[/tex]

Solving for v₁:

[tex]v_1=\sqrt{v_0^2-2gh_1}[/tex]

There is no real solution, because the stone never reaches 25m.

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