Question

An object is projected at u m/s at elevation angle \(\theta\) at a height of 10m. The object hits the ground 100m away. Suppose \(\theta\) = \(tan^{-1}\left(\frac{3}{4}\right)\)

Find the velocity \(u\)

Answer 1

This what I did:
\[Max~ height: \frac{dy}{dt} = -9.8t + \frac{3}{5} u =0 => t = \frac{3u}{49} \]
\[Flight time: 2* t ~to ~reach~ the \max. height => 2*\frac{3u}{49} = \frac{6u}{49} \]
\[x = \left(\frac{4}{5}u\right) t => x = \left(\frac{4}{5}u\right) * \frac{6u}{49} => \frac{24u^{2}}{245} = 100 => u = 31..\theta \]

Answer 2

\(u = 31..\)

Answer 3

Where did i go wrong?

Answer 4

check problem flight line

Answer 5

flight time?

Answer 6

yes

Answer 7

did you take the 2 times the faction

Answer 8

i dont think there is anything wrong with it :/
flight time = 2 * time to reach the max. height

Answer 9

2* (3u/49) => ( 6u/49)

Answer 10

what is the answer?

Answer 11

u = 30

Answer 12

okay so help me to understand your last line and why there are two 4/5c in this line for?

Answer 13

where did that come from? I don't see it?

Answer 14

the first x = (4/5)ut is the question then i subbed the t in which is equal to => x = (4/5)u*(6u/49)

Answer 15

i mean equation not question lol

Answer 16

but you have it twice not once

Answer 17

that is my question where or how did the second 4/5c come into play?

Answer 18

easier to understand now
\[=> x = \left(\frac{4}{5}u\right) * \frac{6u}{49} => \frac{24u^{2}}{245} = 100 => u = 31\]

Answer 19

that is the range
\[=> x = \left(\frac{4}{5}u\right) t\]

Answer 20

please bare with me I am trying to see where you are off is all

Answer 21

ok thanks. i have no idea where i went wrong; it looks right to me

Answer 22

yes you are right, although being off one is odd..

Answer 23

where did i go wrong?

Answer 24

that is why I am asking questions to you nothing more

Answer 25

hey I think the best way know where you went wrong is to start again. May be the mistake is small and can not be seen otherwise

Answer 26

well i did the several times and got the same results that is why i posted here..

Answer 27

are you working from a text book?

Answer 28

nope

Answer 29

where did this problem and answer come from than?

Answer 30

from my notes ; i dont think that its a problem from a textbook

Answer 31

how do know than it's 30 not 31 than?? help to understand I am lost here if not from a text book and only from your notes? just trying to understand this better

Answer 32

idea check your notes again and make sure you copied the answer right. It could be that you are right and the answer you copied is just wrong. just a thought

Answer 33

well; its a problem from my teachers set of notes that he gave to me; which i believe he made it himself so its not from a textbook. And the answer is right it cannot be wrong.

Answer 34

forget it about it. thanks anyway

Answer 35

there is a chance the teacher may have made a mistake and didn't catch it. I have had it happen to me

Answer 36

well the problem is that there is a solution in it which gives the right answer; but there are several ways of solving the problem. that is why i wanted to try another approach; which indeed failed; but i did not know why :/

Answer 37

wait a second how about we take the answer and work backwards now

Answer 38

to see if the answer gives us the start of the problem

Answer 39

i dont need to work backwards.
i just want to know why my approach went wrong

Answer 40

hm, forget about this problem; i will try another approach.
thanks anywy!

Answer 41

no problem I am srry I can not help you :(

Answer 42

CAN I HELP?

Answer 43

yes please!

Answer 44

@sauravshakya