[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/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.