Horizontally Launched projectile landing on curved ground

[bluemyner]

Homework Statement

A watermelon is launched into a ravine horizontally with an initial velocity of 12 m/s. A cross section of the bank has the shape of the bottom half of a parabola, with its vertex at the initial location of the projected watermelon and with the equation y^2 = 16x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?

Relevant Equations

y^2 = 16x

If I'm being completely honest, I have never done a problem where the projectile is being launched over a curved ground so I am not sure where that given equation plays a part in finding my x and y components.

 

[Antarres]

Well, the problem is basically the same as any standard problem with horizontal projectile motion. You need to calculate the trajectory of the projectile, so find y(x) for the projectile, regardless of this surface it is landing on. Then once you've found that, you just search for an intersection between the trajectory and the curve of the ground. That's where it will land.

 

[haruspex]

Homework Statement: A watermelon is launched into a ravine horizontally with an initial velocity of 12 m/s. A cross section of the bank has the shape of the bottom half of a parabola, with its vertex at the initial location of the projected watermelon and with the equation y^2 = 16x, where x and y are measured in meters. What are the x and y coordinates of the melon when it splatters on the bank?
Homework Equations: y^2 = 16x

The description is confusing.
If the vertex is at the bottom and that is the launch point then the projectile will hit the ground immediately.
If the vertex is at the bottom and the launch point somewhere else we need to be told where.
If the vertex is at the top it is an inverted parabola and the equation should have -16 as the factor.
Is there a diagram?

Btw, "Homework Equations" is supposed to be for standard mechanics equations, like Newton's laws, not equations specified as part of the question.

 

[TSny]

The ravine is the lower half of a "sideways" parabola with vertex at the left.

 

[haruspex]

The ravine is the lower half of a "sideways" parabola with vertex at the left.

Doh! y²~x, not y~x². Thanks,

 

[bluemyner]

Well, the problem is basically the same as any standard problem with horizontal projectile motion. You need to calculate the trajectory of the projectile, so find y(x) for the projectile, regardless of this surface it is landing on. Then once you've found that, you just search for an intersection between the trajectory and the curve of the ground. That's where it will land.

I solved for t with the equation:
4(x)^1/2 = 0t + 1/2(9.8)(t)^2
t = (x)^1/4√(4/4.9)
t = .904(x)^1/4

I'm not sure where I should/could go from here.

 

[PeroK]

I solved for t with the equation:
4(x)^1/2 = 0t + 1/2(9.8)(t)^2
t = (x)^1/4√(4/4.9)
t = .904(x)^1/4

I'm not sure where I should/could go from here.

Which direction are you taking as $x$?

Does finding $t$ help?

 

[bluemyner]

Which direction are you taking as $x$?

Does finding $t$ help?

I'm not sure if I totally understand your first question but I'm taking taking the change in y as the equation that was given in the original problem. I figured find t would allow me to somehow find the change in x and then see where the trajectory intersect the parabola but I'm not sure how to do that.

 

[phinds]

Please draw a diagram. You should find it helpful.

 

[PeroK]

I'm not sure if I totally understand your first question but I'm taking taking the change in y as the equation that was given in the original problem. I figured find t would allow me to somehow find the change in x and then see where the trajectory intersect the parabola but I'm not sure how to do that.

I'm assuming you're taking $x$ as horizontal.

Okay, so you've got the x coordinate of the trajectory as a function of $t$?

Isn't that simply $x = v_0 t$?

 

[Antarres]

I solved for t with the equation:
4(x)^1/2 = 0t + 1/2(9.8)(t)^2
t = (x)^1/4√(4/4.9)
t = .904(x)^1/4

I'm not sure where I should/could go from here.

As I said, generally when you're looking for a place of collision, you want to intersect trajectories, if you have the way of getting the trajectories. Trajectory is a line in space along which the particle is traveling over time, so it has the form $y(x)$. You know well that horizontal projectile motion is defined with equations:
$x = v_x t \qquad y = -\frac{gt^2}{2}$
So what you did was that you used this equation for $y$ and you then put it in the equation that describes the landscape $y^2 = 16x$. This makes no sense, as you can see, the equations above describe motion, while $y^2 = 16x$ describes the shape of the ground. So in order to find the trajectory of motion, you want to eliminate the parameter $t$ in the equation, and get the relation between $x$ and $y$. This relation is a curve in space called trajectory, the curve along which the body is moving. Now the question -where- the body will hit the ground has a simple answer. It will hit the ground at the point where it's trajectory touches the ground. So we conclude, we want to find an intersection of the trajectory and the curve that describes the ground $y^2 = 16x$.

That's what I previously said in more detail.

 

[PeroK]

I'm not sure if I totally understand your first question but I'm taking taking the change in y as the equation that was given in the original problem. I figured find t would allow me to somehow find the change in x and then see where the trajectory intersect the parabola but I'm not sure how to do that.

Assuming you have only studied projectile motion onto a flat surface, it might be an idea to look at projectile motion onto an incline.

How would you work out where the projectile lands in that case? You could, for simplicity, take an initially horizontal trajectory and an angle of 45°, say.

That might help you understand the concepts involved.