Question

A daring 510-N swimmer dives off a cliff with a running horizontal leap,what must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge of the bottom, which is 1.75m wide and 9.00m below the top of the cliff.

Answer 1

use simple kinematics

Answer 2

ok
are u familiar with this formula
\[s=ut+\frac{ 1 }{ 2 }g t^2\]

Answer 3

yes sir

Answer 4

whats the first step sir ??

Answer 5

so here initial velocity that is u is equal to zero
so u'll have
\[s=\frac{ 1 }{ 2 }g t^2\]
s=9.00 m that is given
so
\[9=\frac{ 1 }{ 2 }g t^2\rightarrow t=\sqrt{18/g}\]
so \[t=1.35\]
okay so far?

Answer 6

yes i understand now and next ?

Answer 7

by step sir please so i can practice

Answer 8

now it is given that
she leaves the top of the cliff so that she will miss the ledge of the bottom, which is 1.75m wide
so she must travel 1.75 meter
now simply use
\[d=(vecocity)*(time)=v*t\]
so
\[v=\frac{ d }{ t }=\frac{ 1.75 }{ 1.35 }=1.29 m/s\]

Answer 9

got this?

Answer 10

@sidsiddhartha xplained it very well but just to conclude

Answer 11

yeah a diagram is always better :)

Answer 12

yup @sidsiddhartha

Answer 13

Step1
Find the time taken in travelling 9m from the top of the cliff. Suppose u got t seconds

Step2
Find the velocity required to travel 1.75 m in t seconds

Answer 14

@Abhisar i know that sir but i didnt know the formula already

Answer 15

@sidsiddhartha explained everything very nicely....He even did the calculations for you. I hope you know the formulas now !

Answer 16

do u ?

Answer 17

only that is the formula ???

Answer 18

There is no direct formula for this question (if u r expecting that). Two formulas are used here,

\(\large\sf S=ut+\frac{1}{2}gt^2\)
and
Speed = Distance/time

Answer 19

yes sir

Answer 20

sir can you teel the unknown

Answer 21

s=distance or height [9 m in this case]
u=initial speed [0 in this case]
t=time taken [1.35 seconds]
g=acceleration due to gravity [9.8 m/s]

Answer 22

so what 1.29 m/s ?? is that the initial velocity ??

Answer 23

It is horizontal speed by which the diver should run in order to just miss the ledge.

Answer 24

so thats the ans. ???????????

Answer 25

0 is the vertical component of initial velocity and 1.29 is the horizontal component of initial velocity. Yes the answer is 1.29 m/s

Answer 26

Have you learned projectile motion ?

Answer 27

not yet sir but this my ps in projectile motion

Answer 28

\(\underline{\href{//www.physicsclassroom.com/class/vectors/u3l2a.cfm}{\sf \huge This}}\) can be helpful in your learning

Answer 29

thank you sir ,, sir explain everything plsssss

Answer 30

i would hav explained you but it is very long topic and i doubt i have that much time ryt now. I'll urge you to go through it by urself. Mark the doubts and ask me next time. Can you do this ?

Answer 31

mm can you solve it to me sir

Answer 32

sir

Answer 33

Solve what ?

Answer 34

this i post si

Answer 35

but it's already solved !

Answer 36

\(\color{blue}{\text{Originally Posted by}}\) @sidsiddhartha
so here initial velocity that is u is equal to zero
so u'll have
\[s=\frac{ 1 }{ 2 }g t^2\]
s=9.00 m that is given
so
\[9=\frac{ 1 }{ 2 }g t^2\rightarrow t=\sqrt{18/g}\]
so \[t=1.35\]
okay so far?
\(\color{blue}{\text{End of Quote}}\)

Answer 37

\(\color{blue}{\text{Originally Posted by}}\) @sidsiddhartha
now it is given that
she leaves the top of the cliff so that she will miss the ledge of the bottom, which is 1.75m wide
so she must travel 1.75 meter
now simply use
\[d=(vecocity)*(time)=v*t\]
so
\[v=\frac{ d }{ t }=\frac{ 1.75 }{ 1.35 }=1.29 m/s\]
\(\color{blue}{\text{End of Quote}}\)