Understanding Exercise 3 and 8 in Freefall and Motion Law

[Andrei0408]

Homework Statement

I've attached the pictures to this

Relevant Equations

F=ma, dot product

I need help with the exercises attached in the pictures. Basically, exercise 3 is already solved but I need some help understanding every subpoint (for example, at a) how did we get to 0=H -g* tAO/2 , I know it's from the motion law and vA=0, but why is y(t)=0?). And I tried solving ex 8, but I need help at a) and c). I calculated the velocity and the acc but I don't know how I could find the angle between the two (cross product was my guess but I'm not sure), and c) I know that to find the trajectory eq I need to eliminate time, but I don't know for sure how I'm supposed to do that. Even if you don't explain both exercises or not all subpoints, anything is appreciated as I have a test next week and I need to understand. Thank you!

 

[haruspex]

Homework Statement:: I've attached the pictures to this
Relevant Equations:: F=ma, dot product

why is y(t)=0?

You can choose the reference point for having height zero however you like. The author has chosen the ground as illustrated in the diagram, so the initial height is H and height at time of interest (landing) is zero.

Homework Statement:: I've attached the pictures to this
Relevant Equations:: F=ma, dot product

angle between the two (cross product was my guess

That certainly works as a step along tge way, but the dot product is easier.
Given vectors $\vec u$ and $\vec v$ with angle $\theta$ between them, what expressions can you write for $|\vec u.\vec v|$ and $|\vec u\times \vec v|$ in terms of $|\vec u|$, $|\vec v|$ and angle $\theta$?

Homework Statement:: I've attached the pictures to this
Relevant Equations:: F=ma, dot product

to find the trajectory eq I need to eliminate time

You are almost there. You have t equal to a function of x and t equal to a function of y, so...

 

[Andrei0408]

You can choose the reference point for having height zero however you like. The author has chosen the ground as illustrated in the diagram, so the initial height is H and height at time of interest (landing) is zero.

That certainly works as a step along tge way, but the dot product is easier.
Given vectors $\vec u$ and $\vec v$ with angle $\theta$ between them, what expressions can you write for $|\vec u.\vec v|$ and $|\vec u\times \vec v|$ in terms of $|\vec u|$, $|\vec v|$ and angle $\theta$?

You are almost there. You have t equal to a function of x and t equal to a function of y, so...

I mean to say dot product there. But I get v=100t which would give me a weird result. Could you help me a bit more? Also at c) I know I need to eliminate time, so I subtracted the two relations, is this correct?

 

[haruspex]

I get v=100t

No, you got $\vec v.\vec a=100t$. How do you get from there to the angle? Can you answer the question I asked about the dot product in post #2?

at c) I know I need to eliminate time, so I subtracted the two relations, is this correct?

Your answer looks right.

 

[Andrei0408]

You can choose the reference point for having height zero however you like. The author has chosen the ground as illustrated in the diagram, so the initial height is H and height at time of interest (landing) is zero.

That certainly works as a step along tge way, but the dot product is easier.
Given vectors $\vec u$ and $\vec v$ with angle $\theta$ between them, what expressions can you write for $|\vec u.\vec v|$ and $|\vec u\times \vec v|$ in terms of $|\vec u|$, $|\vec v|$ and angle $\theta$?

You are almost there. You have t equal to a function of x and t equal to a function of y, so...

I mean to say dot product there.

No, you got $\vec v.\vec a=100t$. How do you get from there to the angle? Can you answer the question I asked about the dot product in post #2?

Your answer looks right.

Well v.a = |v| * |a| * cosθ and v x a = |v| * |a| * sinθ

 

[haruspex]

I mean to say dot product there.
Well v.a = |v| * |a| * cosθ and v x a = |v| * |a| * sinθ

Roght, so what do you get for cosθ from your dot product?

 

[Andrei0408]

Roght, so what do you get for cosθ from your dot product?

cosθ = v.a / |v| * |a| . The thing was, I couldn't calculate |v| since that would be sqrt(16 + 100t), because there is a t. But now that you've mentioned the cross product, I can calculate v x a with the determinant(it's 40k), and if I calculate v.a/v x a the magnitudes of the vectors v and a reduce and I get cosθ/sinθ = 100t/40k, which is cotθ=100t/40k. So θ=arccot(100t/40k). But even in this form, I can't really get a proper angle. Did I complicate things or is this right?

 

[haruspex]

that would be sqrt(16 + 100t)

t².
The question does not specify a value for t, so it is to be expected that the answer is a function of t.

 

[Andrei0408]

t².
The question does not specify a value for t, so it is to be expected that the answer is a function of t.

Thank you!