How Does Moving at 20sqrt(2) m/s Affect the Projectile's Trajectory Equation?

[ShizukaSm]

Homework Statement

To an observer not in motion in the ground, the trajectory of a projectile is described by:
$y = x - 10^{-3}x^2$(In SI units)
Where it was adopted x = 0 and y = 0 for the origin point. Determine the trajectory equation for an observer that moves at 20sqrt(2) m/s in relation to the ground, in the positive x axis.

(Answer is $y = \frac{5}{3}x - \frac{1}{360}x^2$)

Homework Equations

Only the equations already given.

The Attempt at a Solution

I have no idea whatsoever on how to deal with this problem. I have learned how to deal with relative position problems, and relative velocity problems, but I have no idea how to change the equation themselves and make them reflect that.

 

[hgfhh123]

Look at each motion separately. First of all, what is the x component of the velocity relative to the stationary observer, from your equation? Then, if the observer is moving with 20√2 m/s, what is the x component relative to him? Do the same for the y component.

Think about what that graph is actually telling you about these values.

 

[ShizukaSm]

I apologize, but I do not understand what you mean. Could you maybe rephrase it, in case I am missing something obvious?

I tried taking the inverse function (So I would have the movement in function of x), but still, that didn't help me.

 

[rbrayana123]

Hmm.. if you start with your basic kinematics relationships, you should be able to construct a relationship between y and x involving the initial velocity angle, initial velocity magnitude and the constant acceleration.

However, in this case, you don't know the initial angle, velocity and accelerations. So, it's a matter of working it out the other way.

EDIT: Also you'll probably be working under the assumption that there is only a deceleration of g in the y direction. With all that, you should get the raw materials for which you can answer the question.

 

[Nickie]

As a scientist, it is important to approach problems like this with a clear and logical thought process. Let's break down the given information and see if we can come up with a solution:

1. The given equation describes the trajectory of a projectile from the perspective of an observer on the ground who is not in motion.
2. The equation includes x and y values, which represent the horizontal and vertical positions of the projectile.
3. The origin point is set at x=0 and y=0 for convenience.

Now, let's consider what would happen if we introduce an observer who is moving at 20sqrt(2) m/s in the positive x axis. This means that this observer is moving along the trajectory of the projectile, but at a faster speed. How would this affect the equation?

We can start by considering the x values. Since the observer is moving at a constant speed, we can assume that the x values will be larger than those of the stationary observer. We can express this as x' = x + vt, where x' is the new x value, x is the original x value, v is the speed of the observer, and t is time.

Next, we can consider the y values. Since the observer is moving at a constant speed, the y values will also be larger than those of the stationary observer. However, we also need to take into account the effect of gravity, which will cause the projectile to follow a parabolic trajectory. This means that the y value will decrease as the x value increases. We can express this as y' = y - 10^{-3}x'^2, where y' is the new y value and x' is the new x value.

Now, we can substitute x' and y' into the original equation and solve for y:

y' = x' - 10^{-3}x'^2
y - 10^{-3}(x + vt)^2 = x + vt - 10^{-3}(x + vt)^2
y = x + vt - 10^{-3}(x^2 + 2xvt + v^2t^2)
y = x + vt - 10^{-3}x^2 - 2vtx - 10^{-3}v^2t^2

We can simplify this by factoring out an x from the last two terms:

y = x(1 - 10^{-3}x - 2