Random projectile motion problems.

[Jebus_Chris]

1. Trajectory of a projectile is modeled by
$$r(t) = x(t)i+y(t)j$$
$$x(t) =at^2+bt$$
$$x(t) =ct+d$$
It asks for the displacement between t = 1 and t=3 s.
And I got
$$x(t)=8a+2b$$
$$y(t)=2c$$
If it is asking for the displacement would the answer be$$x+y$$ or $$\sqrt{x^2+y^2}$$?2. The ball falls down from height h (above incline) on incline with angle 45. Calculate how far from striking point, ball will hit incline again if collision is perfectly elastic.

So the velocity it gains from falling would be $$\sqrt{2gh}$$. That would be the initial velocity of when it reflects at 0 degrees. Then you could create the projectile motion equations for it.
$$x=\sqrt{2gh}t$$
$$y=\frac{1}{2}gt^2$$
It will land when the x and y distances are equal. So after replacing the y with x, solving for t in terms of x, and substituting I got x=4h=y. So then the displacement would be $$\sqrt{8}h$$. That right?
Now I originally did this as the ball reflecting at 45(><) and i got $$x=hcos^2\theta (tan^2\theta - 1)$$ That gives you zero but you know that it would eventually hit the ground, why is that?

 

[anathema]

Hi there,

For the first question, the displacement between t=1 and t=3 seconds would be given by the formula \sqrt{(x(3) - x(1))^2 + (y(3) - y(1))^2}. So the answer would be \sqrt{(8a+6b)^2 + (2c)^2}.

For the second question, I believe you have made a mistake in your calculations. The initial velocity gained from falling would be \sqrt{2gh}sin(45) = \sqrt{gh}. And the projectile motion equations would be:

x = \sqrt{gh}cos(45) t
y = \sqrt{gh}sin(45) t - \frac{1}{2}gt^2

To find the distance from the striking point to where the ball hits the incline again, we need to set x = y and solve for t. This gives us t = \frac{2\sqrt{gh}}{g} = 2\sqrt{\frac{h}{g}}. Substituting this value of t into the equation for x, we get x = \sqrt{gh}cos(45) 2\sqrt{\frac{h}{g}} = 2h. So the displacement would be 2h.

I hope this helps clarify your doubts. Let me know if you have any further questions.

 

[tomcue]

For the first problem, the answer would be the magnitude of the displacement, so it would be the square root of the sum of the squares of the x and y displacements. In other words, it would be \sqrt{(8a+2b)^2 + (2c)^2}.

For the second problem, your approach and final answer are correct. The reason why the ball would eventually hit the ground even though the equation gives a zero value is because the equation assumes a perfect reflection at the incline, which is not realistic. In reality, the ball would lose some energy and not reach the same height as the initial drop, causing it to eventually hit the ground.