(Fun?) kinematic problems

[Tetra]

Just three questions that I want to clarify about the meanings for.

 

Q1: A northern wind (blowing north to south) is blowing at 70m/s. If a jet is flying at 280m/s and the pilot wanted to fly due west, what heading should the plane take? What is the new groundspeed for the plane?

 

Ok, so I Just want to check: this question is asking for what angle the plane should be flying at (somewhere in the NW direction) so that, with the northern wind, it would be going exactly west? And then I find the ground speed too, yes?

 

 

Q2: Mr. Smith throws a bar at with a speed of 30m/s at an angle from the horizontal that gives maximum range. His student is sitting on a desk located 1.5m above ground level ready to catch the bar.

 

a)What are the horizontal and vertical components of the acceleration vector at maximum height? What about velocity?

b) What is the max. height the bar can reach?

 

For a), I;m not sure what the difference between the acceleration and velocity would be for the components. Aren;t they the same in this question? And is it asking for them at the instantaneuos moment at maximum height?

 

And for b), the max height would be 1.5, right, because the student is at that hieght to catch it? Or should we just assume there is no one there to stop it?

[Cap'n Refsmmat]

Q1: Yes, that sounds right.

 

Q2: Why would the acceleration and velocity vectors be the same? For example, a bar in free flight above earth will be constantly accelerating downwards due to gravity, whereas its velocity will change according to that acceleration. The maximum height wouldn't be 1.5m, either -- Mr. Smith throws the bar upward, so it'll go higher.

[Tetra]

So if the bar is at its highest, then it wouldn't be propelled by anything anymore, thus only gravity acts on it. (so 9.8m/s^2). And because there is no change in the position of the object at the max, then velocity would be...0m/s?

 

As for the horzontal, a horizontal projection (component) would have 0 acceleration, while it's velocity is still moving at 30m/s. Is that right? Because even at the max height, it's still moving forward at the constant speed (assuming there is no air resistance or anything).

[Cap'n Refsmmat]

Everything's right but the 30m/s number. Remember, he throws it at an angle, so only a portion of its velocity is in the horizontal component.

[Tetra]

Yeah, that's what I was wondering. I was considering constructing a right triangle to visually show the thing process, then solve with trig. The hypoteneuse would be the 30m/s, and the hieght would have been the 1.5m height of the student.

 

But we can't do that can we, because we don't necessarily know that the student is at the max height, or farther. Butif we can't then we don't really have any information to work. on. How could we find the height when we know only two vertical variables (acceleration and some height?)

[Cap'n Refsmmat]

You know the angle of part of the triangle, because it's the angle the bar is thrown at.

 

Well, you know it's the angle that gives maximum range. You can use it as a variable and solve for it, I think.