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What does the average net force landing exercise mat?

Mass 65kg high jumper to clear only 2.13m tall and drops visit Straight down to 0.30 m thick mat. 1. What is the average net force During the visit is unknown 0.18m high jumper in compressed Landing exercise mat? 2. Why force the jumper to be highly effective if Compress 5cm landed in the sand?

Answer - "(1) None -7,546 (2) -54,328 No down, no longer has the speed to give you lower power Did. Then you find the acceleration, Newton's, in essence: the second. of m = 65kg H = 2.13 m g = 9.81 m / s of the Vi ^ 2 = 0.0 m / s and (1) D = 0.18 m compression of one trillion First The impact of the acceleration of gravity, the speed to solve. Vf = SQRT (Office of ^ 2 + 2ah) Vf = SQRT of ((0.0 m a / m s) ^ 2 + [2 * a (9.81 m / s in ^ 2) * (2.13 m)]) Vf = SQRT (41.7906 미터 ^ 2 / s in ^ 2) Vf = 6.46 / s, so the initial rate is now about Using acceleration to solve a given distance is zero, Rate the final]. = [Vf ^ Vi ^ 2 of - 2.2 D = [(0.0 m / s) to the ^ 2 - (6.46 m / s) ^ as 2] / [2 * (to 0.18 m)] = [-41.791 m ^ 2 / s ^ 2 of] / [0.36 m] ^ 2 Note the negative Symbols / m = -116. You will need it. Newton's second Forced to use search. Group F = m * F and = (65kg) * (-116 m / s in ^ 2) F one trillion = -7,546 N, except to note the symbol. You will need it. (1) Group D = 5cm for the metric units in terms of putting in the same He said it was doing the speed velocity = 0.05 m will affect the final acceleration The distance around a given speed, is still the same initial Is 6.46 right now / more than one m of the s, is zero =. [Vf ^ 2 of - Vi ^ 2] / 2, D = [(/ s of the 0.0 m) ^ 2 - (/ s of the 6.46 m) ^ 2] / [2 * (0.05 m)] = [a - 41.791 M ^ 2 / s in ^ 2] / [0.05 m login negative note 2] = -836 ^ two m / s,. You It should be. Newton's second search is using force. Group F = m * F a = (65kg) * (-836 m / s in ^ 2) F = -54,328 N symbols refer to one trillion excluded. You will need it.

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[ApnaCommunity.net] Sajan Re Jhoot Mat Bolo - 28th june 10 pt1.wmv


 

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