Min Plane Height to Drop Bomb in Hollow: 335m

[blue_eyes_88]

Homework Statement

What is the smallest height for a projectile to hit the target?

Relevant Equations

Kinematic equations

A plane is flying 80km/h in horizontal direction and it has to drop a bomb into 30m wide and 30m deep hollow. What is the smallest possible height for the plane to fly above hollow if the bomb successfully hits the bottom?

I made a mistake somewhere but not sure where... the correct result is 335m

My solving so far:

 

[berkeman]

Welcome to PF.

It kind of looks like you should have subtracted the 30m hole depth instead of adding it. The altitude of the plane is in reference to the surface of the ground, not the bottom of the hole.

I think you approached the question correctly, but I did not go through your work in detail to check it (yet).

 

[blue_eyes_88]

Welcome to PF.

It kind of looks like you should have subtracted the 30m hole depth instead of adding it. The altitude of the plane is in reference to the surface of the ground, not the bottom of the hole.

I think you approached the question correctly, but I did not go through your work in detail to check it (yet).

Thank you for a warm welcome
Even if I subtract 30m, then my result is 365.5m. Still figuring out where did those 60m came from.

 

[berkeman]

But if you subtract the 30m from 365...

I'll try to look through your math in a bit. I think you approached it the way I would have, by having the bomb just barely clear the near lip of the hole and hit the bottom middle of the hole. That gives you two points on the parabolic arc down, and you know the horizontal speed at both of those points.

 

[blue_eyes_88]

But if you subtract the 30m from 365...

I'll try to look through your math in a bit. I think you approached it the way I would have, by having the bomb just barely clear the near lip of the hole and hit the bottom middle of the hole. That gives you two points on the parabolic arc down, and you know the horizontal speed at both of those points.

That's right. Then the smallest possible altitude has to come from time t which we can get from w/2.

 

[TSny]

Looks like your symbols $v_1$ and $v_2$ represent the vertical components of the velocities. Is that right?

So, $v_1$ is the vertical component of velocity of the bomb after it has fallen a height $h_1$?

Shouldn't the 2 that is circled in orange actually be a 4? Otherwise, your work looks good to me.

What value did you use for $v_0$ in m/s?

I'm not getting anything close to the given answer of 335 m for the altitude of the plane. I get about 84 m for your $h_1$. I could be making some errors.

 

[nasu]

The limiting case is not when the bomb hits the middle point of the hole. You can go even lower and hit off center.

 

[berkeman]

The limiting case is not when the bomb hits the middle point of the hole. You can go even lower and hit off center.

Very interesting insight. So as long as you clear the near lip of the hole, the minimum altitude for the drop may not result in the bomb hitting the middle of the hole. I did not think of that case; thank you.

Homework Statement:: What is the smallest height for a projectile to hit the target?
Relevant Equations:: Kinematic equations

if the bomb successfully hits the bottom?

But the problem statement does mention "bottom", so that's why we were assuming the middle/bottom of the hole...

 

[DaveC426913]

"That's impossible! Even with a computer!"

"I used to bullseye Womp-rats in my T-16..."

 

[nasu]

Why would you assume this shape for the hole? Why not a cylindrical one, with flat bottom?

 

[DaveC426913]

Why would you assume this shape for the hole? Why not a cylindrical one, with flat bottom?

That's a good point. A cylindrical hole doubles the horizontal window from 15 to 30m.

 

[blue_eyes_88]

Why would you assume this shape for the hole? Why not a cylindrical one, with flat bottom?

True, good idea. Didn’t even considered it.

 

[blue_eyes_88]

Looks like your symbols $v_1$ and $v_2$ represent the vertical components of the velocities. Is that right?

View attachment 298533

So, $v_1$ is the vertical component of velocity of the bomb after it has fallen a height $h_1$?

View attachment 298532
Shouldn't the 2 that is circled in orange actually be a 4? Otherwise, your work looks good to me.

What value did you use for $v_0$ in m/s?

I'm not getting anything close to the given answer of 335 m for the altitude of the plane. I get about 84 m for your $h_1$. I could be making some errors.

The factor 1/2 is correct. I checked my solution process and there is no factor in front of g^2*t^2 so when dividing by 2 we get 1/2.

 

[blue_eyes_88]

Thank you guys for your help.

It’s clear to me now that my solving process is solid, only have to check now for different parameters (cylinder, maybe they made extra typo for initial condition?).

I appreciate your help!

 

[TSny]

The factor 1/2 is correct. I checked my solution process and there is no factor in front of g^2*t^2 so when dividing by 2 we get 1/2.

But it looks to me that substituting for $t$ using $t = \large \frac{w}{2v_0}$ brings in another factor of 2 in the denominator. Maybe I'm overlooking something.

Also, I don't see how you got 85.5 m/s when substituting values here

If I use w = d = 30 m, g = 10 m/s², and V₀ = 80 km/h = 22.2 m/s, I get about 38 m/s for V₁. If the 2 in the denominator of the second term should actually be 4, then I get V₁ to be about 41 m/s.

Anyway, your approach looks valid to me for the way in which you are interpreting the question.

 

[blue_eyes_88]

OMG, TSny.

You just made me realize how negligent I am when solving problems. No wonder why I am so conscious about myself and never sure if it's wright or wrong. I'm really grateful to you :) I know on what I have to work on (God, I feel so stupid for not putting a ratio in correctly).

Went through the problem again and got 38 m/s ----> h1 = 72.2 m.

Maybe for the official solution there was meant other condition, but I'm happy with this solution.

 

[kuruman]

The problem says that the hole is 30 m deep by 30 m wide. Projectile motion is two-dimensional so I would say that the projectile just barely clears the lip of the hole then travels horizontally inside the hole for 30 m while it drops vertically another 30 m. That is, I imagine a square with the top corner connected to the bottom diagonally-across-corner by the parabolic trajectory of the projectile.

The horizontal component of the velocity is known and the time of flight inside the hole can be found quite easily. One needs to find the vertical velocity component at the point of entry which would determine the height from which the bomb was released. This interpretation gave me a drop height less than 30 m. Forum rules bar me from being more specific.