Question

The range of a projectile thrown horizontally is said to be dependent to the following quantities 'R'~f(v. H, g, m). Derive an expression for R in terms of v, H, g and m.

(I got R = V^2/g but I know if I am just correct because I don't really know the formula for the range of projectiles, can somebody verify?) Thanks!

Answer 1

@hartnn , can you help me?

Answer 2

The solution should depends on H. Do you put in your kinematic equations the initial height as H?

Answer 3

Yes. But it appear like this in the end.
Btw, can you start it? So I can verify if I was in a right path.

Answer 4

for x:
x = vt

so we have to find the flight time t.

for y:
0 = h - 0.5gt^2

now you can find t using y equation
and plug it into x equation and have R.

Answer 5

indeed
x = vsqrt(2h/g)

using dimensional analysis you can assume that
the range will depend on v g and h but not on m since nothing will cancel the mass units.

Answer 6

Exact.

Answer 7

Is the solution you made is dimensional analysis approach? @Coolsector

Answer 8

no

Answer 9

But how can we do it as dimensional analysis approach?

Answer 10

using dimensional analysis we would not know that there is a "2" in the square root

Answer 11

Can you show me how to start?

Answer 12

well i think knowing the answer make me cheat here :
units of distance [m] , time [s] .
we want g, h, and v in the expression:
x ~ v^(a) * g^(b) * h^(c) = [m/s]^a * [m/s^2]^b * [m]^c = [m]

so
a+2b = 0
b = -a/2
and
a+b+c = 1
c = 1-a/2

maybe i miss something ( im not really use to dimensional analysis)
but i think we should make another assumption about a (the power of v)
if we take it to be 1 we get the correct powers

Answer 13

assuming a =1 though is reasonable due to x=vt in such throw

Answer 14

Thanks for telling we can assume a=1
In my solution, I did
R ~ v^w h^x g^y m^z
[R] = L
[v] = LT^-1
[H] = L
[g] = LT^-2
[m] = M

Therefore, K= (LT^-1)^W L^x (LT^-2)^y M^z
L = L^(w+x+y)T^(-w-2y)(M^z)

z=0
w = -2y
w+x+y = 1
Letting w = 1, I get
y = -1/2
x = 1/2
Therefore,
R ~ VH^1/2g^-1/2
or
\[R ~ \frac{ v \sqrt{H} }{ \sqrt{g} }\]
which is
\[R = V \sqrt{\frac{ H }{ g }}\]
HAHAHA! I got it. Thanks, but how letting w= 1 reasonable??

Answer 15

as i said:
"assuming a =1 though is reasonable due to x=vt in such throw"
i hope it is correct and not "cheating"
anyway you cant have an equal sign .. using dimensions you can only have proportional..

Answer 16

Okay. I think it's true anyway. Thanks! :)

Answer 17

about the proportional :
as you can see we didnt get the 2 under the square root as we got in the "x = vsqrt(2h/g)" which is the correct full answer

Answer 18

@luigi0210 @nincompoop
can I really substitute 1 in the w in my solution here?

Answer 19

@psymon