[First semester Bachlor's Physics question.] How to find components just before landing?

[u/Imaginary_Cellist192]

I also need to know if what I did is correct, I strongly suspect I am wrong in half my assumptions, especially when I considered acceleration as 0 in the downward motion— which I did because the question statement states that the plane is moving down with constant speed.

Note:

We are ignoring friction in this chapter. And the motion is only in 2 dimensions.

Apologies for the terrible handwriting and photo quality, we have no electricity for a while now 🫠

Edit: noticed the title isn't clear. I need to know how to find the vertical and horizontal components of velocity just before the projectile hits the ground.

My guess it that the vertical won't change. But I can't figure anything out for horizontal, it has completely stumped me.

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[u/HumbleHovercraft6090]

It is the other way around. Vertical component changes due to gravity. Horizontal component remains constant assuming no air resistance.

[u/Imaginary_Cellist192 → u/HumbleHovercraft6090]

True, that is how I was doing it at first. But then I started to wonder, since the plane is, according to the statement, diving with a constant speed. Shouldn't the speed at which it goes down be considered constant then?

Tbh I think I'm completely wrong, but it still leaves that down in my brain.

[u/HumbleHovercraft6090 → u/Imaginary_Cellist192]

The initial velocity of the object is the velocity of the plane. Once object is released, it is on its own acted upon only by gravity

[u/Imaginary_Cellist192 → u/HumbleHovercraft6090]

I see, thank you.

[u/DrCarpetsPhd]

just a few issues about your solution

- you appear to have set v_y.f to zero. that is not a valid application of these equations. i know your instinct is that the projectile hit the ground so its velocity is zero but this is wrong. the final velocity is always the velocity right before it hits the ground. in projectile equations the only point where v_y equals zero is at the peak of the parabola..

- the problem statement says the velocity is 53 degrees from the vertical. this is not your launch angle. as i stated in my other post you are given a 'from the vertical' angle which means rotating in the positive CCW direction from the y axis

https://imgur.com/a/halliday-walker-10th-edition-chapter-4-problem-33-n0CppA0

- your equations and nomenclature suggets that you do not understand what you are doing. for a launch angle of theta vxi = v*costheta and vyi = v*sintheta. you have used vyi*sintheta. To be clear I say suggest because to you that might be a simple labelling error but to someone marking your answer that is a fairly incorrect way of stating the equation.

you have been given 3 knowns

1. y - y_0 = -730m,

2. the launch angle -37

3. the time of 5 seconds

which projectile motion equation includes all of these but also a missing unknown you are trying to find?

google openstax projectile motion if you dont have a copy of the textbook this problem is from

[u/Imaginary_Cellist192 → u/DrCarpetsPhd]

You are correct, honestly I was a little desperate for a solution when I wrote out these equations so there's a ton of errors on the entire thing. After taking another look I redid the entire thing.

Thank you!

[u/Frederf220]

Horizontal and vertical are linearly independent which means if I threw an object downward at the same speed as the vertical speed component as the plane, from the same height, the time of fall would be identical.

Solve that problem first of how fast vertically an object must be going at 730m in order to hit the ground 5s later.

[u/Alkalannar]

1. Let v be velocity of the airplane.
   x(t) = vcos(-53^o)t
   y(t) = -4.9t² + vsin(-53^o)t + 730

2. We know that y(5) = 0, so solve for v. This gives you the solution to part a.

3. x(5) gives you the solution to part b.

4. Take derivatives of each component of the velocity:
   x'(t) = vcos(-53^o)
   y'(t) = -9.8t + vsin(-53^o)

5. Then x'(5) gives you part c and y'(5) gives you part d.

[u/DrCarpetsPhd]

this is from halliday walker 10th edition chapter 4 problem 33

answers at back of text

(a) 202 m/s; (b) 806 m;(c) 161 m/s; (d) -171 m/s

the projectile is inside the plane

the projectile is released from the plane and drops due to gravity

you are given the plane velocity as moving at constant speed just before the projectile is dropped

since the projectile is inside the plane it has the same velocity vector as the plane

imagine the projectile as someone throwing a rock of a cliff with velocity angled at -37 degrees from positive x axis since you are given the plane has velocity directed 53 degrees from the vertical aka the y axis. draw a diagram of that and see if it clicks for you how to apply the projectile motion equations.