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Imshaniago
Answered

a marble is thrown horizontally...

b.) FORMULA:
[tex]vfy = - \sqrt{2gdy} [/tex]
[tex]vr = \sqrt{vfy { }^{2} + vfx {}^{2} } \\ [/tex]
SOLUTION:
[tex]vfy = \sqrt{(2)(9.8ms {}^{2})(1.08m) } \\ vfy = \sqrt{21.168ms} \\ vfy = - 4.6ms[/tex]
[tex]vr = \sqrt{(4.6ms) {}^{2} + (1.50ms) {}^{2}) } \\ vr = \sqrt{21.16m {}^{2} s {}^{2} +2.25m {}^{2}s {}^{2} } \\ vr = \sqrt{23.41m {}^{2} s {}^{2} } \\ vr = 4.84ms[/tex]
therefore, the velocity of the marble before it hits the floor is 4.84m/s

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Jhomellviernesgo

a marble is thrown horizontally...

b.) FORMULA:

[tex]vfy = - \sqrt{2gdy} [/tex]

[tex]vr = \sqrt{vfy { }^{2} + vfx {}^{2} } \\ [/tex]

SOLUTION:

[tex]vfy = \sqrt{(2)(9.8ms {}^{2})(1.08m) } \\ vfy = \sqrt{21.168ms} \\ vfy = - 4.6ms[/tex]

[tex]vr = \sqrt{(4.6ms) {}^{2} + (1.50ms) {}^{2}) } \\ vr = \sqrt{21.16m {}^{2} s {}^{2} +2.25m {}^{2}s {}^{2} } \\ vr = \sqrt{23.41m {}^{2} s {}^{2} } \\ vr = 4.84ms[/tex]

therefore, the velocity of the marble before it hits the floor is 4.84m/s

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