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Physics 21 Online
OpenStudy (anonymous):

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.

OpenStudy (anonymous):

what is "h" ?

OpenStudy (anonymous):

Use the energy method. Decrease in potential energy= gain in kinetic energy mg(H-h) = 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.

OpenStudy (anonymous):

Hope its clear

OpenStudy (anonymous):

(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

OpenStudy (anonymous):

isn't "d" the distance covered after he jumped?

OpenStudy (anonymous):

i think D is ski jump's length

OpenStudy (anonymous):

i drew above do you solve it?

OpenStudy (anonymous):

D should be the range i feel.

OpenStudy (anonymous):

Let us take D to be the range. Now, decrease in potential energy would be mg(H-h) as he came down a height of H-h So, mg(H-h) = 1/2 mv^2 v=(2g(H-h))^1/2

OpenStudy (anonymous):

question want H did you get it?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

so it is...?

OpenStudy (anonymous):

approximately 15.026

OpenStudy (anonymous):

how did you get time of descent

OpenStudy (anonymous):

t= {2h/g}^1/2

OpenStudy (anonymous):

this t is time of H time of descent is: 2*t

OpenStudy (anonymous):

how did you take v=u in two formula? mg(H-h) = 1/2 mv^2 R= u. t

OpenStudy (anonymous):

D= v. t where v= [2g(H-h)]^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

OpenStudy (anonymous):

u is velocity of body when it's arrive to ground but vis velocity in top of path

OpenStudy (anonymous):

no i think you use 2*t in formula for D

OpenStudy (anonymous):

hence d is all of horizontally path

OpenStudy (anonymous):

|dw:1331643793186:dw|

OpenStudy (anonymous):

"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

OpenStudy (anonymous):

Yes "D" is the horizontal distance covered.

OpenStudy (anonymous):

did you see my diagram ? H is max height that man arrive . i didn't understand your fig

OpenStudy (anonymous):

i mean that how did you yeild to u=v

OpenStudy (anonymous):

When 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.

OpenStudy (anonymous):

mistyped u for v

OpenStudy (anonymous):

am i clear?

OpenStudy (anonymous):

yes you'r right

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