Question

HELP!!!!!A particle is dropped from a height,h above ground!
If it covers 1/3rd of its distance in its last sec,find the time of flight!!!!!!HELP

Answer 1

@thomaster @Kainui @karatechopper @jhonyy9 Help Guys!!

Answer 2

So what things do you know? The acceleration due to gravity and several equations relating position, velocity, acceleration, and time. Draw a picture and label it and try to use the equations you have along with some algebra to solve for what you need.

Answer 3

It goes 1/3 the way in a second? is that what it means?

Answer 4

and then u find the total flight?

Answer 5

Thats the question.....1/3rd of its total distance in the last second!!

Answer 6

Yes

Answer 7

I have done it in a way and found an answer.Dunno if its r8!!

Answer 8

Shall I post it?

Answer 9

I need another way to find the answer,easier if any!!Dunno if this's r8!!

Answer 10

Any Idea @karatechopper ??

Answer 11

sorry I don't quite know.. :/

Answer 12

K...tnx anyways!!

Answer 13

so, we can write the equation for the last second:

0=1/3 -v1 - g/2
where i plugged 1

Answer 14

plugged t =1

Answer 15

now we can find v1 (the velocity 1 second before the particle hits the ground)

Answer 16

v1 = 1/3 - g/2 (it will be positive since i took a negative sign)

Answer 17

Why take t=1??

Answer 18

i just want to find the initial velocity before the last second

Answer 19

what is g in those units ?

Answer 20

32.17?

Answer 21

9.8

Answer 22

but you gave data in yards ? :O

Answer 23

M not gtin ya

Answer 24

oh sorryyy grr

Answer 25

No ....There's no hint of units of distance!!

Answer 26

sorry sorry

Answer 27

Np...cntu!!

Answer 28

Did I do it in the correct way??.Hav posted my work above!!

Answer 29

i might slipped somewhere though

Answer 30

wait

Answer 31

anyway im not focused but this is how i would do it.. im sure there are mistakes here

Answer 32

NP coolsector

Answer 33

Tnx for Ur Valiant effort!!

Answer 34

tnx @Coolsector ....Guess no one can help me with this!!

Answer 35

tnx.... @karatechopper !!

Answer 36

tnx.......@Kanui

Answer 37

tnx.......@Kainui

Answer 38

@Kainui

Answer 39

@Coolsector ...Okay if I close this post??

Answer 40

ok so using
v^2 = v0^2 + 2g(y-y0)
we find the velocity before the last part of flight (that h/3) :
v^2 = 2g(h/3 - h)
v^2 = -4hg/3
v = sqrt(-4hg/3)

also
v/g=t

now the last part: (1 second so t=1)
0 = h/3 + v + g/2
0 = h/3 + sqrt(-4hg/3) + g/2

h=3sqrt(6)g -15g/2

Answer 41

ok i dont mind
i keep on doing mistakes anyway lol

Answer 42

but this is the way to the solution anyway for sure

Answer 43

ok

Answer 44

I think I'll let it remain open for a while....Sumone's out there who will know the answer!!

Answer 45

@Coolsector

Answer 46

My thanks to @Coolsector for the derivation of the velocity \(\frac{2}{3}\) down! \(v^2=v_0^2+2\ a\ \Delta d\)
\(=v_{top}^2+2\ g\ \left(\frac{2}{3}h\right)=2\ g\ \left(\frac{2}{3}h\right)\)


Now tow find the entire height:
\(d=d_0+v_{\tfrac{2}{3}}\ t+\frac{1}{2}\ a\ t^2\)
\(\qquad\Downarrow\)
\(h=\frac{2}{3}h+v_{\tfrac{2}{3}}\ t+\frac{1}{2}\ g\ t^2\)

Now you can solve for \(h\).

That's what I've gotten to so far. Do you have any questions @cambrige ?

Answer 47

\(d=v_i\ \Delta t+\frac{1}{2}\ a\ \Delta t^2\)
\(=h=(0)\Delta t+\frac{1}{2}\ g\ \Delta t^2\)
So, now you can solve for the time interval for the entire fall, \(\Delta t\).

Answer 48

Let the total height be \(h\) and the total time of flight be \(t\).
Using \(\displaystyle d = d_{0}+v_{0}t +\frac{1}{2}at^2\) for the entire motion, \(d_0 = 0\) and \(v_0 = 0\), So
\(\displaystyle h=\frac{1}{2}gt^2\) ...... (1)

Now consider the motion for the time \(t-1\) and use the same equation. Here \(d = 2h/3\)
\(\displaystyle \frac{2h}{3} = 0 + 0\cdot (t-1) + \frac{1}{2}g(t-1)^2\) Substitute the value of \(h\) obtained in (1)
\(\displaystyle \frac{1}{3}\cancel{g}t^2 = \frac{1}{2}\cancel{g}(t-1)^2\) Now take squareroot on both sides

\(\displaystyle \frac{\pm t}{\sqrt 3} = \frac{t-1}{\sqrt 2}\) or \(\displaystyle t = 3 \pm \sqrt 6\)

Since the time of flight has to be atleast greater than 1 sec (since it covers h/3 in the last second), so the only answer possible is \(t = 3+\sqrt 6\)

Answer 49

@FoolAroundMath :)

Answer 50

So My ans is right,I guess...

Answer 51

No @theEric....No more Q's....Tnx for the help!!

Answer 52

Tnx to @FoolAroundMath