Air resistance physics (1 Viewer)

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Hi guys, got a question and been trying to solve it for 2 days.

the question is:

In the absence of air resistance with what angle to a horizontal ground must an object be thrown from ground level so that it ravels a distance before landing equal to twice the maximum height ?

I know that i've got to work with simaltaneius equations, tried all the linear equations but can't get to an answer. Any one can show me the steps to arrive to the answer pls ?

Thank you in advance
 

rsk

159
4
Set up an equation to calculate the maximum vertical height. Use vf^2 = v0^2 + 2 a s

At this point the projectile has travelled how far horizontally?

Can you equate these now to solve for the angle?


Edited......haven't tried this yet but you might need x = v0t + 0.5 a t^2 for the vertical instead, to be able to solve...
 
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rsk said:
Set up an equation to calculate the maximum vertical height. Use vf^2 = v0^2 + 2 a s

At this point the projectile has travelled how far horizontally?

Can you equate these now to solve for the angle?


Edited......haven't tried this yet but you might need x = v0t + 0.5 a t^2 for the vertical instead, to be able to solve...

i tried that mate still didn't manage to. i'll show you what i did.

Vertical:
s=ut+0.5at^2
s= usinΘ*t + 0.5(-10)(t^2)
s=usinΘt+5t^2

Horz:
s=ut
2s=ucosΘ*2t (2t because maximum height =t, and it takes double that to go down again, 2s because double the distance)

then

tried to substitute but can't come to a conclusin.

I also tried s=ut+0.5at^2 but still can't :S
 
Zeno's Paradox said:
This might help: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra14

[tex]\frac{h}{R} = \frac{\tan \theta}{4} = 1/2 [/tex]​

Then, you may find theta.

Note: You should not memorize this formula, but understand how to deduce it.
I'll try to use that. Thnaks to that i know that the answer should be 63.43494882292201 :rolleyes:
 

rsk

159
4
Yes, sorry, I'm still trying too and can't get it.

Each time I try I either end up with lots of unknowns, or else everything cancels out to leave something trivial.

WIll post back when I get somewhere!
 
rsk said:
Yes, sorry, I'm still trying too and can't get it.

Each time I try I either end up with lots of unknowns, or else everything cancels out to leave something trivial.

WIll post back when I get somewhere!
thx mate, i've spent at least 4 hours trying to work them out, but always got a lot of unknowns. Mainly the time keeps me away from solving it,

So i thought i would get out an equation for time, but then i got like a cos^-1 of 0 which is not possible.
 

rsk

159
4
got it

In the vertical, use v^2 = u^2 + 2as to find an expression for the max height.

I got h = (usina)^2/19.6 ------------------(1)

(a of course is the angle, I don't have alpha or theta!)

Then use x = ut + 0.5at^2 to find an expression for T, the time for the object to land (ie vertial displacement = 0).

I got T = usina/4.9

Then in the horizontal, the obhect travels 2h in this time T

so 2h= ucosaT

or h = (ucosa)T/2 ---------------(2)

I then equated these 2 expression for h, and after a bit of rearranging got

sina/cosa = 19.6/9.8

or tana = 2



Try it and check there are no stupid errors......
 
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144
0
You have that the equation of motion in 2 dimensions is

[tex] \mathbf{r}(t) = (v_0\cos(\theta)t) \mathbf {\mathrm{i}} + (-gt^2/2 + v_0\sin(\theta)t)\mathbf{\mathrm{j}} [/tex]

Where [itex] \theta [/itex] is your firing angel to horizontal ground. From here you can find an expression for how long the projectile travels horizontally. Because what is the vertical component of the vector [itex] \mathbf{r} [/itex] equals to, when the projectile hits the ground? When you know that, you can solve the vertical equation for the time. You can then substitute that expression for the time in the horizontal component of the vector to find out how far along the horizontal the projectile traveled.

To the find the maximum heigh above the ground level, you have to use the assumption that the vertical component of you velocity vector [itex] \mathbf{\dot{r}}(t) [/itex] is equal to zero at the maximum height. Solve that equation for the time, and when you substitute that expression for the time into your vertical component of the vector [itex] \mathbf{r}(t) [/itex] you find the height, and for this time it will be te maximum height.
 

rsk

159
4
Given that what I ended up with is basically tana = 2g/g, i wouldn't be surprised if there was a quicker way to do it.....
 
rsk said:
got it

In the vertical, use v^2 = u^2 + 2as to find an expression for the max height.

I got h = (usina)^2/19.6 ------------------(1)

(a of course is the angle, I don't have alpha or theta!)

Then use x = ut + 0.5at^2 to find an expression for T, the time for the object to land (ie vertial displacement = 0).

I got T = usina/4.9

Then in the horizontal, the obhect travels 2h in this time T

so 2h= ucosaT

or h = (ucosa)T/2 ---------------(2)

I then equated these 2 expression for h, and after a bit of rearranging got

sina/cosa = 19.6/9.8

or tana = 2



Try it and check there are no stupid errors......
Understand it. But can get how you got

"Then use x = ut + 0.5at^2 to find an expression for T, the time for the object to land (ie vertial displacement = 0).

I got T = usina/4.9"

could you please give me the steps to arrive to that usina/4.9
 

rsk

159
4
at this time the vertical diplacement is =0 - it's back at the level it started.

so 0 = ut + 0.5 a t^2

so 0 = usina T - 4.9 T^2

so usina T = 4.9 T^2

cancel T from each side

u sina = 4.9T

so T = u sin a / 4.9
 
thanks rsk for explaining everything :)

worked everything, and it's 100% correct.
 

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