# I don't have a clue how to find the coordinates

• lioric
In summary: In my view, the vertex of the parabola is the point where the melon is launched, and the melon lands on the bank at the point where the parabola intersects the bank. The "road" is irrelevant to the solution of the problem.In summary, the problem involves finding the coordinates of the point where a parabola, representing the trajectory of a melon, intersects with a given bank. The melon is launched from the vertex of the parabola and lands on the bank at the point of intersection. The road is not relevant to the solution.
lioric

## Homework Statement

http://i277.photobucket.com/albums/kk63/lioricsilver/Untitled_zps7cf41f04.png

## Homework Equations

y2=(16)x is the equation on the question

## The Attempt at a Solution

I have got any clue. I do know how to solve motion in 2d. But since time is not given or the height is not given I don't know how to apply any of what i know to this question.

But my hypothesis is that since the hyperbola has an equation given i should use it to find minnimum of the curve. But that's all i got. Sorry but i found all attempts a dead end befor i begin because there is so less values to work with

#### Attachments

• Untitled.png
47.2 KB · Views: 514
What condition must hold when the melon hits the bank?

If you were to find the path of the melon when it is launched horizontally with the given initial speed, could you calculate the x and y values of the trajectory, ignoring the shape of the bank?

lioric said:
But since time is not given or the height is not given I don't know how to apply any of what i know to this question.
The problem is not particularly well-worded. The vertex of the parabola must be the point where the melon takes off, as shown in the picture, not the edge of the road, as stated in the problem description. So the origin of the system is also as shown in the picture, and therefore the initial height is zero. (Of course, in general the actual location of the origin is arbitrary, but here it makes sense to locate it at the vertex of the parabola.)

You have, or can get, an equation for the trajectory of the melon, and an equation for the stream bed. What must be true, as SteamKing asked, when the melon meets the ground?

Ok after analyzing the data i got here is what i know and don't knowObject horizontal stuff

constant velocity V=10m/s
Acceleration a=0
Time t=?
Distance s=?

Vertical Stuff

Initial Velocity U=0
Final velocity V=?
Acceleration = 9.81m/s2(10m/s2)
Time t=?
Distance s=?

But i do know that the time taken for both is the same

Know where do i go from here?

Start by writing down the relevant equations. Those would be the kinematics equations that relate position, velocity, acceleration, and time.

I did try v=u+at for the vertical velocity

so v=0+10 x t
v=10t

also for vertical i tried s=ut+1/2at2
so s=0t+1/210t2
s=0+5t2
s=5t2

also i tried v2=u2+2as
so v2=0+2 x 10 x s
v2=20s

what do i do now?

I also mixed up the first 2 equations

v=10t and s=5t2 by substituting in the place of t and got
v=10 x √ s / √ 5

square all to get rid of root and got

v2=100 x s/5

substitute this with 3rd equation v2=20s

and got 400s2= 100s/5

so make subject s

and by now i feel faint

some one please give me hand please. I need some sort of guidance for this

that equation that is given holds the final key

y2=16x

so i guess i need to put distance in x and y even in terms of s and solve

Plz tell me if this could work

And i couldn't understand what steamking asked in post 2

Well the question says the melon splatters on the bank

What can we say about the x and y coordinates of the melon and the x and y coordinates of the bank when the melon hits the bank?

tms said:
The problem is not particularly well-worded. The vertex of the parabola must be the point where the melon takes off, as shown in the picture, not the edge of the road, as stated in the problem description. So the origin of the system is also as shown in the picture, and therefore the initial height is zero. (Of course, in general the actual location of the origin is arbitrary, but here it makes sense to locate it at the vertex of the parabola.)
...
It looks to me as if the problem is worded just fine. It's the art work that doesn't fit the problem too well.
lioric said:
I did try v=u+at for the vertical velocity

so v=0+10 x t
v=10t
No. The horizontal component of v is a constant: vx = 10 m/s .

The x coordinate of the melon, as a function of time is
x = vx∙t

x = 10∙t .​

also for vertical i tried s=ut+1/2at2
so s=0t+1/210t2
s=0+5t2
s=5t2
Note: use the X2 icon for super scripts.

You are almost correct regarding the vertical position of the melon. (Of course the melon's vertical coordinate goes down from zero, not up.)

So the y component of the melon is
y = -5∙t2
.

Solve x = 10∙t for t and plug that into the equation for y, to find the path taken by the melon.

You now have two simultaneous equations in x and y to solve .

Thank you very.
I have got it by now the only thing that was bothering me was the equation of the parabola.
But i got it now

Could i post all my working and final answer so that you could tell me if it is right or wrong?

lioric said:
Could i post all my working and final answer so that you could tell me if it is right or wrong?
Yes, that is perfectly fine.

Distance in x axis
s=ut+1/2at2

s=? U=10 t=? a=0

∴ s=10t

Distance in y axis
s=ut+1/2at2

s=? U=0 t=? a=9.8

s=9.8/2 t2

t in x-axis and t in y-axis is same

substitute distance in x and y of y2= 16x

y2= 16x

(9.8/2 t2)2 = 16 x 10t

24.01t4=160t

make subject t

24.01t4/t= 160

24.01t3=160

t3=160/24.01

t3= 6.664

t=3√6.664

t=1.88s

----------------------------------

substitute t= 1.88s in equations in x and y axis

s=10t

10x 1.88

=18.81

s=9.8/2 t2

9.8/2 (1.88)2

x= 18.81
y=17.35

is this correct?

I gave everything in very detail so you can tell me where i went wrong.

Last edited:
lioric said:
Distance in x axis
s=ut+1/2at2

s=? U=10 t=? a=0

∴ s=10t
That's for the x coordinate of the melon.

Therefore, x = 10t for the melon.

Distance in y axis
s=ut+1/2at2

s=? U=0 t=? a=9.8

s=9.8/2 t2

That's for the y coordinate of the melon, except that it should be -9.8/2 t2.

Therefore, y = -9.8/2 t2 for the melon.

...

Solve x = 10t for t.

Substitute that into y = -9.8/2 t2.

That will give y in terms of t, for the melon's trajectory.

Take this result together with the equation for the river bank.

This gives two equations in two unknowns. Solve for x any y. (There are two solutions.)

## 1. How do I find the coordinates of a location?

Finding the coordinates of a location can be done in several ways. The most common method is by using a GPS device or a smartphone with a GPS feature. You can also use online tools or maps to search for the coordinates of a specific location.

## 2. What are coordinates and why are they important?

Coordinates are a set of numbers that represent a specific location on the Earth's surface. They are used to pinpoint a precise location and are important for navigation, mapping, and other scientific and engineering purposes.

## 3. How do I convert between different coordinate systems?

Converting between different coordinate systems can be done using online tools or software specifically designed for this purpose. You will need to know the type of coordinate system you are converting from and to, and follow the instructions provided by the tool or software.

## 4. Can I find the coordinates of a location without using technology?

Yes, you can find the coordinates of a location without using technology. You can use a map and a compass to determine the coordinates of a location by triangulation. This method involves drawing lines on a map from known points to the location you want to find, and using the angle and distance measurements to calculate the coordinates.

## 5. How accurate are coordinates?

The accuracy of coordinates depends on the method used to determine them. GPS devices and online tools typically provide coordinates with an accuracy of a few meters. However, manual methods like triangulation may have a lower accuracy. The accuracy can also be affected by external factors such as interference and atmospheric conditions.

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