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Let us use v2 = u2 + 2 a S

v = 0, S = 0.27 m and a = g = -9.8 m/s2

Plugging and solving for u we get, u = 2.3 m/s

Hence take-off velocity is 2.3 m/s

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āˆ™ 11y ago
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āˆ™ 1mo ago

To determine the take-off velocity of a jump of 0.27 m, you would need to know additional information, such as the angle at which the jump is launched, acceleration due to gravity, and the height of the jump. These variables are necessary to calculate the initial velocity required to achieve a jump height of 0.27 m.

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Q: Which is the take-off velocity of a jump of 0.27 m?
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