Question

Using uH and time of flight 2T, find an expression for horizontal distance, 5m.

Answer 1

Could you possibly explain where this "uH" comes from and what it means?

In general, distance = rate * time.

If distance in your situation is 5m, and the time elapsed is 2T,

then distance = 5m = ( rate )* ( 2T ).

Is it possible that you mean:\[x _{h}=v _{h}*t\] ... ??

Answer 2

My previous questions were, make an expression for the vertical and horizontal components of initial velocity uV &uH. my answer was vertical uV=U sin theta and uH=U cos theta. Also, assuming no resistance determine an expression for the time T when projectile reaches its maximum height. my answer was: T=Uv/g.

Hope that helps

Answer 3

Unfortunately, your post here was taken out of context, and I therefore had almost no information to go by. Also, it's quite unusual to use "u" to represent velocity.

I'd be happy to consider helping with this problem if you'd present the whole problem at once, along with any illustrations and instructions. Could you possibly take one or more screen shots to upload (using Attach File)?

Answer 4

this what i have answered so far

Answer 5

Just for info - when I learned the ballistic equations we used
u = initial velocity
v = final velocity
s = distance (displacement)

thus
\[v ^{2}=u ^{2}+2as\]
or
\[s=ut+\frac{ at ^{2} }{ 2 }\]

etc

Answer 6

so does g for gravity not come into the final equation?

Answer 7

g is acceleration of gravity

the equations above work for all uniform acceleration

So substitute g for a if working with gravity (although this is only true if working with small objects, close to earth's surface)

Answer 8

so how do i feed a distance of 5m into that equation?

Answer 9

There is no accelerationin the horizontal direction

s=ut

u = uh
s=5m

Answer 10

Not sure that it is 5metres

Your time of flight is 2T (where T is time to max height)
You really need to post the question in its full original form if you want help with the whole thing....

Answer 11

the full question is the question at the top and 4th reply down has an attachment with my work so far on 2 previous similar questions

Answer 12

There is no mention in your document of a value of 5m, or of any ACTUAL velocity

So - does the question mean

'Give an expression for calculating the time for a horizontal distance travelled = 5m'

If so - use the equation I gave in my answer 2 above (s=ut) there is no acceleration in horizontal direction.

(for a cannon this is likely to be a very short time, which is why I queried your question)

Answer 13

I apologize for the 5. i believe i read it wrong, here is exactly how it is written on my paper:
Using UH and the total time of flight, 2T, ie. the time to rise vertically plus the
time to fall to ground, find an expression for the horizontal distance, S m,
covered by the projectile

Answer 14

OK

So - I have given you the informayion above:

There is no horizontal acceleration so in general terms

s=ut

put in your values for u (horizontal velocity) and t (total time of flight)

Answer 15

thanks very much, appreciate all the help!