Hard Physics Problem, can someone help? - physics of a belly flop
I've tried, but I can not seem to understand ..... Please help. Thank you!
A relative of your stomach bug from a height of 2.50 meters (ouch!) And he continues to move after falling to 0,500 meters under water. Its mass is 62.5 kg. (a) What is the speed when it hits the water? Ignore air resistance. (b) How large is the degree of dynamism, when in the water, and if there any more? (c) What was the rate of acceleration in the pool? Let us assume that is constant. (d) How long is the water before it no longer moves? (e) What was the rate of average net force exerted on him after he hit the water until it stops?
Wednesday, February 17, 2010
Physics Of A Belly Flop Hard Physics Problem, Can Someone Help?
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a) v ^ 2 = u ^ 2 +2 as a formula. a is the acceleration of gravity. Therefore, v ^ 2 = (0) ^ 2 + 2 (10) (2.5) = 50 m / s
ReplyDeleteb) You must at the time of the first situation. s = ut + 1 / 2 A ^ 2 = 0.5 (0) t + 1 / 2 (10 t) ^ 2. t = 0.316.
F = m (VU) / t = 62.5 (50-0) / 0.316 = 9889.2
c) = (VU) / t = (50-0) / 0.316 = 158.228m / s ^ -2
Not sure about other problems. Hope this helps.:)