Question

In planet x the formula T=au gives the time of flight of a particle thrown vertically up with a velocity u in that planent the formula followed for final velocity of a particle dropped from rest from a height h is?

Answer 1

What is \(T\)?
What is \(a\)?

Answer 2

if T is time a -acceleration and u is velocity then this formula is not correct, due to its dimensions. I'd write this formula in another way
T = u/a

Answer 3

@open_study1

Answer 4

^Yup. So there are issues in this question's posing.

Answer 5

it is said it take place in other planent

Answer 6

Regardless of what planet this takes place in, unless it is outside of our universe and part of another universe in the multiverse, the laws of Physics still apply. Hence, our critique still stands.

Answer 7

So... Any idea what \(T\) and \(a\) are?

Answer 8

KInematics does not depend on particular planets characteristics. The laws of physics are supposed to be the same everywhere, though it is not proved, in my point of view ))).

Answer 9

T is the time of flight and a is acceleration in question it is aid t=au we have to use it and find v

Answer 10

You have said that \(u\) is the velocity.

Is this incorrect? What does \(u\) mean?

Answer 11

i mean velocity

Answer 12

It is obviously a typo. Ok let's assume that
t = u/a and find the formula for the velocity.

Answer 13

T = total time of flight

Answer 14

@ArchiePhysics

Answer 15

@shivam_bhalla plzz help

Answer 16

\(v=v_0+at^2\)
If \(h\) is the height the particle is dropped from, we have that \(h=\frac{1}{2}at^2\) (via distance formula)
Thus,
\(v=v_0+2h\)
\(v_0=0\)
\(v=2h\)

Anyone care to correct me? This seems deceptively simple.

Answer 17

v can't be equal to h because of dimensions

Answer 18

i will give the options 2wholeroot h/a wholeroot h/a wholeroot 2 h/a wholeroot h/2a

Answer 19

@ArchiePhysics plzz just see whether u r getting by using this T=au

Answer 20

\[v_f^2=v_i^2+2ad\]
\[v_i=0\]
\[d=h\]
\[v_f^2=2ah\]
\[v_f=\sqrt{2ah}\]

Answer 21

There is no such option lol @Limitless

Answer 22

(I'm aware it doesn't match the given answers.)

Answer 23

thats i told to use T=ua

Answer 24

@ArchiePhysics wat u say

Answer 25

Even if admit that T = au
my answer is different.

Answer 26

None of those answers make sense to me. It is impossible for \(t=au\) where \(a\) and \(u\) are acceleration and velocity, respectively. Making such a statement renders our formulas nonsensical.

Answer 27

Plz @Limitless u just use this a see that u get

Answer 28

Just to be sure what you mean by wholeroot?

Answer 29

\[\sqrt{h/a}\]

Answer 30

@ArchiePhysics Like this

Answer 31

ok

Answer 32

Maybe I am wrong, but none of these answers work out properly. (Check them dimensionally.)

Answer 33

here is my solution. i haven't checked through for mistakes yet

T=au

assuming constant acceleration (negligible air resistance)

V = U - At

for upwards flight:
V = 0
U= u
t = T/2

0 = u - AT/2


A = 2u/T = 2/a

now at a height h we have mAh gravitational potential energy, and this equals kinetic energy at bottom

mAh = (1/2)mv^2

\[v = \sqrt{2Ah} = \sqrt{\frac{4h}{a}} = 2 \sqrt{\frac{h}{a}}\]

Answer 34

What does \(A\) represent as compared to, say, \(a\)?

Answer 35

A is the acceleration due to gravity, i just assumed a was some constant or whatever..

Answer 36

T = au
is not dimensionally consistent which would explain why final answers aren't if we assume something about the dimensions of a

Answer 37

@eigenschmeigen you are absolutely right if a is some constant then it is easy-peasy problem!

Answer 38

What do you posit that \(u\) is? And \(T\)?

Openstudy said they were velocity and time, respectively.

Answer 39

i assumed they were velocity and time

Answer 40

Then, if \(T=au\) where \(a\) is some number, \(u\) is velocity, and \(T\) is time, we are still dimensionally inconsistent.

Answer 41

a is a constant with dimension of
[a] = [t/u]=[s^2/m]

Answer 42

yeaah

Answer 43

That makes sense, @ArchiePhysics. I am not certain, but perhaps @eigenschmeigen did solve this after all. :)

Answer 44

Hey Shakir, where are you? You've got the answer.
@openstudy1

Answer 45

Personal curiosity: What could \(a\) be called? I don't immediately recall what the inverse of acceleration is.

Answer 46

hmm interesting. i haven't encountered it as an actual quality we describe an object having. at rest or uniform motion it would be undefined

Answer 47

@mahmit2012

Answer 48

i didnt understand can some one help me...

Answer 49

i did nt understand the last step written by @eigenschmeigen