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what is "h" ?
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.
Hope its clear
(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
isn't "d" the distance covered after he jumped?
i think D is ski jump's length
i drew above do you solve it?
D should be the range i feel.
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
question want H did you get it?
so it is...?
how did you get time of descent
this t is time of H time of descent is: 2*t
how did you take v=u in two formula? mg(H-h) = 1/2 mv^2 R= u. t
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
u is velocity of body when it's arrive to ground but vis velocity in top of path
no i think you use 2*t in formula for D
hence d is all of horizontally path
"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
Yes "D" is the horizontal distance covered.
did you see my diagram ? H is max height that man arrive . i didn't understand your fig
i mean that how did you yeild to u=v
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.
mistyped u for v
am i clear?
yes you'r right