A 75.0 kg stunt man jumps from a balcony and falls 20.5 m before colliding with a pile of mattresses. If the mattresses are compressed 0.900 m before he is brought to rest, what is the average force exerted by the mattresses on the stuntman?
_____ N
2007-02-23 07:36:40 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
the velocity of man just as he reaches the mattresses be v. Then v^2 = 0+2*9.8*20.5=
(v^2=u^2+2*a*s)
Let's assume average force to be F.
retardation on man due to mattresses will be (F/75) m/s^2
0= v^2+(-2*(F/75)*0.9)
therefore, 2*9.8*20.5= 2*(F/75)*0.9
hence, F= 9.8*20.5*75/0.9= 16741.66 N
2007-02-23 08:03:04 · answer #1 · answered by Pulkit 1 · 0⤊ 0⤋
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2016-12-14 04:05:06 · answer #2 · answered by ? 4 · 0⤊ 0⤋
75 * 9.81 Newton ( at the moment the matrasses are full compressed I assume ideal compression no losses, so afterwards the matrass launches the stunt man back to the spot where he started,
2007-02-23 07:41:30 · answer #3 · answered by gjmb1960 7 · 0⤊ 1⤋
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2007-02-24 08:41:02 · answer #4 · answered by Big Baller 1 · 1⤊ 0⤋
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2007-02-23 07:46:58 · answer #5 · answered by beyonce 1 · 0⤊ 0⤋
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2007-02-23 07:46:38 · answer #6 · answered by steve h 2 · 0⤊ 0⤋