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Ghella [55]
3 years ago
7

A pole vaulter takes 1.3 seconds to fall from peak height to the landing mat. What is his vertical velocity at contact with the

mat?
Physics
1 answer:
julia-pushkina [17]3 years ago
6 0

12.75 \frac{m}{s}

Assuming vertical acceleration of 9.81 \frac{m}{s^2}, the speed after x seconds of falling is 9.81x \to 9.81(1.3) = 12.75

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Answer:

the   tension in the string an instant before it broke = 34 N

Explanation:

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\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \  m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\

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goldfiish [28.3K]

Answer:

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