Question

a body is thrown vertically up with a speed of 10 m/s from the top of a wall of height 3.45m the height from which another body should be dropped simultaneously such that two bodies reach the ground in same time (g=10m/s2) explanation??

Answer 1

I don't understand why you are using 10m/s^2 for gravity.
It should be 9.8

Answer 2

IT IS THERE IN QUESTION (G=10M/S2

Answer 3

that dosent really matter much romero infact it simplifies the question though i still havent figured it out :P

Answer 4

Well it's not realistic. using 9.8 vs using 10 is a BIG difference that amounts to a great error.

Answer 5

Ok so you have a wall that is 3.45 m right?

Answer 6

Are both bodies released at the same time?

Answer 7

well if both bodies start falling at the same time then all objects weather light or heavy reach the earth at same time as it dosent depend on mass

Answer 8

it does nt matter about the mass

Answer 9

options are 40m 24.5m 6.9m 20m and the answer is 24.5 but i dont know the steps to do it

Answer 10

Oh easy just calculate the distance the first body travels vertically up before the force of gravity stops. AND the mass does matter because you need it to determine how much you need over time to make it stop.

Answer 11

can u give the steps plzz

Answer 12

I can't I don't know what the mass is.

Answer 13

Actually you don't.

Answer 14

Do you know the kinematic equations?

Answer 15

no it dosent matter according s=vt or t=s/v
the equation for acceleration due to gravity would be S=vf_vi+at where vf is the final speed vi is the initial speed

Answer 16

I know I just want to know if the asker knew the what the kinematic equations were. - _ -

Answer 17

yes i know the equations

Answer 18

plz solve it for me.....

Answer 19

shameer have uou written the complete numerical because it dosent say what should be calculared if you left something out please include

Answer 20

*calculated

Answer 21

the question is complete one i rechecked it...

Answer 22

Well the kinematic equations involved 5 unknowns (actually 6 but we join to of them)
t= time
vi= initial velocity
vf=final velocity
a= acceleration
d= distance traveled.

How many of these do we know?

Answer 23

acceleration and initial velocity are given

Answer 24

how about vf=vi+gt

Answer 25

we need to find the distance

Answer 26

i am a little confused about the question when is the second body thrown downwards?

Answer 27

two bodies are thrown and dropped at the same time

Answer 28

@warz I understand the questions and the answer please let me finish and you can follow along if you want to understand it as well.

Answer 29

vf2_vi2=2gS

Answer 30

That would giv u the distance

Answer 31

d= (vi + vf)/2 * t
this also gives you the distrance.

Answer 32

romero can u give the solved steps and answer plzzz

Answer 33

Great so we know the initial velocity and the acceleration and we want the distance traveled.

Like I said before the force of gravity will bring the body to stop at some point so actually know the final velocity.

Answer 34

yes V=0

Answer 35

with
vi
vf
a

and solving for d
we use

\[vf^2= vi^2 + 2*a*d\]

Answer 36

Solve for d

Answer 37

\[(vf^2 - vi^2)/2a= d\]

Answer 38

plug in the given values
vi=10m/s
vf=0m/s
a=10m/s^2

Answer 39

then d= 5m

Answer 40

yes note also that we have a negative vi but because gravity acts against the direct of the object acceleration will also be negative so that cancels the first negative and it makes it positive.

Answer 41

ok then what to do???

Answer 42

the final answer should be 24.5m

Answer 43

Add it to the height of the cliff.

Answer 44

so d=8.45 m

Answer 45

next what to do??

Answer 46

That should be your answer.

Answer 47

but thats not the answer

Answer 48

it is 24.5m but i dont know how to solve

Answer 49

Then you wrote the question wrong. It's worded poorly and it might have confused me. I can see why I might have gotten the answer wrong. @warz also questioned if you wrote the question properly so you might want to go back and see if you wrote it right.

Answer 50

a body is thrown vertically up with a speed of 10 m/s from the top of a wall of height 3.45m the height from which another body should be dropped simultaneously such that two bodies reach the ground in same time (g=10m/s2)

Answer 51

i checked the question once more it does nt have any mistake...

Answer 52

That "sentence" has a lot of grammar mistakes.

Answer 53

it is the question from my book..

Answer 54

any help needed?

Answer 55

or am i late?

Answer 56

he is right the question demands the explanation to how both bodies hit the ground at the same tim

Answer 57

aravind help us out

Answer 58

u r in at right time

Answer 59

np you are not late you could help by explaining the question to us

Answer 60

This can mean a lot of things. It can mean when the velocity of the body thrown up goes to zero and at the same time we have the second body dropped at the same time at the same location vf=0

Answer 61

That's what I solved for.

Answer 62

:) well basically the qn says that a 2 bodies
1-dropeed
2-thrown up
they have same final velocities
we need to find height from which second body was thrown

Answer 63

lemme wrk out in a paper then i will giv the answer

Answer 64

ok

Answer 65

We know what it is :3.45m

Answer 66

@AravindG can you explain me one thing about the question please is the second body thrown downwards at the same time the first is thrown upwards

Answer 67

yes @warz

Answer 68

ya i gt it!!

Answer 69

The equation of kinematics for motion with constant acceleration in current case is
\[h = h _{0}+v _{0}t-(g \times t^{2})/2\] [1]
we put - sign before \[(g \times t^{2})/2\] because in our problem we have chosen upwards direction as positive, so acceleration due to gravity is negative because it Earth pulls down all objects on it.
First body travels firstly upwards then downwards so in order to find entire time we must solve two equations:
1) v0=gt => t = v0/g or t1=10/10= 1s for upwards motion
from first equation we find
\[h = 2.2+10t-(10 \times 1^{2})/2=7.2 m\] it is the maximum height first object reaches
2) \[h=(g \times t ^{2}) /2\] => \[t=\sqrt{( 2 \times h)/g}=\sqrt{( 2 \times 7.2)/10}=1.2 s\]
so \[t _{total}=1s+1.2s=2.2\] for first object.
We apply the same procedure for second object
and get
1) v0=gt => t = v0/g or t1=10/10= 1s for upwards motion
from first equation we find
\[h = 1.05+10t-(10 \times 1^{2})/2=6.05 m\] it is the maximum height second object reaches

2) \[h=(g \times t ^{2}) /2\] => \[t=\sqrt{( 2 \times h)/g}=\sqrt{( 2 \times 6.05)/10}=1.1 s\]
so \[t _{total}=1s+1.1s=2.2\] for first object.

Now the ratio of appropriate times is
t1/t2=2.2/2.1=22/21
the answer is A.

Answer 70

That's what I asked at first and you never said yes.

Answer 71

A?

Answer 72

Oh sorry for the second object \[t _{total}=1s+1.1s=2.1s\]

Answer 73

@shameer1 are u satisfied with that answer ??else i will show my work

Answer 74

@AravindG Let me get this straight. We know the first object is dropped at 3.45m . We want to know the height the second object is thrown up so that both objects hit the ground at the same time right?

Answer 75

hahaha shameer u have got eve one scratching there heads i always used to keep a hand book for such buggers :D

Answer 76

@Romero the second object is dropped down

Answer 77

@shameer1 ,pls use use punctuations wherever necessary else the reader might get the qn wrong

Answer 78

@AravindG
We know the first object is thrown at 3.45m . We want to know the height the second object is dropped so that both objects hit the ground at the same time right?

Better?

Answer 79

hey aravind plz show ur work

Answer 80

ya exactly

Answer 81

@Romero u should include the word simultaneously

Answer 82

the answer must be 24.5m

Answer 83

k i will show my work

Answer 84

the second object would be a chicken feather i guess thats why it floated cause of air friction :D

Answer 85

@Shameer have you changed the text of question?

Answer 86

see no one understood what the question asked for. lol worded poorly.

Answer 87

hey that was just such a good explanation that a body thrown fa greater hight would have greater speed

Answer 88

vf2_vi2=2gh greater the value of h greater the speed bingo

Answer 89

well first we calculate time taken by the first body ,

consider the upward motion of it
well the time taken can be found out using
\[\large v=u+at\]
\[\large 0=10-10t\]
which gives t=1s
the distance travelled upward will be
\[H=10t-\frac{1}{2}\times5\times1^2\]
\[=5m\]

now so at topmost point height of first body will be \[\large 3.45+5=8.45\]
\[\large8.45=\frac{1}{2}\times10\times t^2\]

solving which we get t=1.3s

Answer 90

now for both bodies the time is same 1.3s

Answer 91

wait a sec the height of second body i am getting is 8.45m

Answer 92

both bodies actually travel 2.3 s
you forgot about the 1s

Answer 93

oops ths romero

Answer 94

thx

Answer 95

@Romero exactly

Answer 96

we need to find distance

Answer 97

ya i gt 26.45