Question

A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the bottom of the cliff. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at a constant acceleration a and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , a , h , and the gravitational acceleration g .

Answer 1

\[v^2 = u^2 + 2as \]

Answer 2

yes but i am having problem find s
v=sqrt(2dg)
but how to combine h to find s

Answer 3

s = ut + 1/2 a t ^2 = h
use this equation for vertical analysis and gravity for acceleration

and use that answer and previous equation for horizontal analysis

Answer 4

but how can relate s with the above two variable

Answer 5

no idea ...Sorry

Answer 6

i got this one but did you got the last question and the archer one?

Answer 7

could u please tell me how and i have not tried archer one yet but this is awfully similar to that monkey problem

Answer 8

break the trip up into to parts ,
running, and falling

start with the second part solve the vertical equation for t

Answer 9

I'm sorry @tini12 , I'm not really good in physics.. :(

Answer 10

sir i am having problem finding t
i have broken it uo

Answer 11

\[y(t)=h-\tfrac12gt^2\]

Answer 12

first calculate at what speed he jumps off the cliff..

Answer 13

it is aa parabolic trajectory u cant solve that way

Answer 14

find v using v^2=u^2+2as then t using 2eq of motion ...then put s=v*t

Answer 15

@tini 12

before the parabollic trajectory, he runs in a st line.. correct?
so first find what is the speed with which he jumps off the cliff

Answer 16

v=sqrt(2ad)

Answer 17

yeas

Answer 18

good..

now consider the parabollic motion with that initial speed..

Answer 19

yes you got v, now find t and then multply then

Answer 20

h=-1/2gt^2

Answer 21

no minus sign

Answer 22

t=sqrt(2h/g)

Answer 23

thnks

Answer 24

did you got any other sum?

Answer 25

what is \[x(t)\] for the falling part?

Answer 26

yes "throwing projectile"

Answer 27

x(t)=v_0t nevertheless i found the answer mr.rhaukas

Answer 28

yeah you have v_0=v above and t

Answer 29

can any 1 please tell me the answer

Answer 30

2*sqrt((a*d*h)/g)
@satty622

Answer 31

sorry @tini12 for not sticking around last night, i meant to come back and got sidetracked by family stuff
did you finally understand how to do it?

Answer 32

yes

Answer 33

if not, to sum it all in one post:

analyzing backwards, we know that to clear distance "s" (horizontal) in time "t", he must have been travelling at a certain horizontal speed (call it x meters/second).
(Horizontal) Equation to use - \[s = ut + \frac{ 1 }{ 2 }at^2\]
in this case: u = x = horizontal speed
now we know he's not accelerating (horizontally) after he runs off the cliff, so in this case a = 0, and u = x; therefore the equation to use shortens to become => \[s = xt + 0 \rightarrow \rightarrow \rightarrow s= xt\]
now the time "t" he spent in the air is only determined by the height of the cliff "h"... and nothing else.(acceleration "g" due to gravity is constant, and initial vertical speed =0)

(Vertical) Equation to use - \[s = ut + \frac{ 1 }{ 2 }at^2\]
in this case:
u = initial vertical speed = 0 meters/second
a = vertical acceleration = gravity "g" (note = 9.8 m/s^2)
s = vertical distance he falls = height of cliff "h"
therefore the equation becomes => \[h = 0 + \frac{ 1 }{ 2 }gt^2 \rightarrow \rightarrow \rightarrow h = \frac{ 1 }{ 2 }gt^2\]
which, rearranged in terms of time (t) becomes => \[h = \frac{ 1 }{ 2 }gt^2 \rightarrow \rightarrow \rightarrow t = \sqrt{\frac{ 2h }{ g }} \]

Finally, the runners speed (x) is determined by the length of the run-up distance "d":
(Horizontal) Equation to use - \[v = u + at\] note: this time "t" is seperate and different from the other previous t value (which equal each other), so lets call this time "y" (seconds)
now: v = final velocity = x m/s
u = "at rest" therefore = 0 m/s
a = runners horizontal acceleration = a
t = time he runs before he reaches the edge = y seconds

so: \[v = u + at \rightarrow \rightarrow \rightarrow x = 0 + ay \rightarrow \rightarrow \rightarrow x = ay\]

Answer 34

rearranged into one usable equation:
\[s= xt\] where \[x^2 = u^2 + 2ad\] and \[t = \sqrt{\frac{ 2h }{ g }} \]
so
\[s= \sqrt{2ad} \times \sqrt{\frac{ 2h }{ g }}\]