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anonymous
 4 years ago
A skier (m=57.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 6.20 m and D = 13.4 m, find H.
anonymous
 4 years ago
A skier (m=57.0 kg) starts sliding down from the top of a ski jump with negligible friction and takes off horizontally. If h = 6.20 m and D = 13.4 m, find H.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Use the energy method. Decrease in potential energy= gain in kinetic energy mg(Hh) = 1/2 mv^2 here we get the velocity. After taking off, the range is given by R= u. t where "t" is the time of descent and 'u' is the velocity.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(1/2)mv^2=mgD so v=sqrt(2gD) v on above is v0 for next part of ask (when skier leave ski jump) y=(v0^2/2*g) so H=y+h

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0isn't "d" the distance covered after he jumped?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i think D is ski jump's length

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i drew above do you solve it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0D should be the range i feel.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let us take D to be the range. Now, decrease in potential energy would be mg(Hh) as he came down a height of Hh So, mg(Hh) = 1/2 mv^2 v=(2g(Hh))^1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0question want H did you get it?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you get time of descent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this t is time of H time of descent is: 2*t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0how did you take v=u in two formula? mg(Hh) = 1/2 mv^2 R= u. t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0D= v. t where v= [2g(Hh)]^1/2 and t= [2h/g]^1/2 "t' is the time of descent from "h" to the ground. I typed u instead of v

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u is velocity of body when it's arrive to ground but vis velocity in top of path

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i think you use 2*t in formula for D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hence d is all of horizontally path

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1331643793186:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0"v' is the velocity of the man before taking off. We are not concerned about the velocity which he will have while landing on the ground

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes "D" is the horizontal distance covered.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you see my diagram ? H is max height that man arrive . i didn't understand your fig

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i mean that how did you yeild to u=v

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When a body takes off horizontally, the height from which it took off will be the maximum height. i took that the person slides from a height "H" and after reaching a height "h', he takes off and covers a horizontal distance 'D". I mistyped u or v.
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