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When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high? 16 V 8 V 4V 2V squareroot 2V

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adelekeademolu

Answer:

[tex]\sqrt{2}V[/tex]

Explanation:

Velocity at maximum height is 0 m/s. With initial velocity, V, height h and acceleration of gravity, g,

[tex]v_f^2 = v_i^2+2as[/tex]

[tex]0^2 = V^2 -2gh[/tex]

g is negative because the motion is against gravity.

[tex]V = \sqrt{2gh}[/tex]

If the maximum height were doubled, then we have

[tex]V_1=\sqrt{2g\times2h} = \sqrt{2}\sqrt{2gh} = \sqrt{2}V[/tex]

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