How Much Explosive Force is Needed to Overcome Drag and Elevate a Car?

[vbottas]

Homework Statement

A 25000kg armored car, with dimensions, 6m by 3m, coefficient of drag of 0.30. Assume density of air to be 1.3 kg/m^3. Was launched 1.72 meters in the air, 35 degrees above the horizontal at its center of gravity due to an explosion by a mine. . Assuming there was drag force pushing the car down during the explosion, how many pounds of explosives was needed to lift the car to this height and angle.

Edit: Missing piece of information. The armour car was gunning at its top speed of 70mph. Foot down to the floor.

Relevant Equations

Drag force formula, potential energy formula. F=ma possibly?

Hello! Apologies if I'm missing some information but here is my thinking so far. Had the drag force not been there, the potential energy would equal the amount of energy the explosives have to disipate to get the car to that height. However, since drag force is included it makes it more complicated and not sure where to go from here. Thank you everyone!

 

[vbottas]

I have a diagram of this drawn out. Let me know if you would like me to attach it for more clarification!

 

[haruspex]

I assume you are using the drag formula quoted at https://en.wikipedia.org/wiki/Drag_coefficient.

Can you write the horizontal and vertical F=ma differential equations?

 

[vbottas]

I assume you are using the drag formula quoted at https://en.wikipedia.org/wiki/Drag_coefficient.

Can you write the horizontal and vertical F=ma differential equations?

In my mind, In the vertical direction, there is 3 forces. Force from the blast, FDrag and, F of gravity.
Fblast-Fdrag-mg=ma
will the blast require horziontal components? I was origionally thinking no...

Edit: Now that I think of it, does gravity need to be added? Becuase had there been no drag force, we wouldn't even have to worry about gravity as the potentional energy= energy from the explosive.

 

[haruspex]

Let me know if you would like me to attach it for more clarification!

The only thing I need clarifying is the angle. Is that the launch angle, or the angle of the line from the launch point to its highest point?

In my mind, In the vertical direction, there is 3 forces. Force from the blast, FDrag and, F of gravity.

No, the blast is a reminder one-off impulse. Once off the ground, there are just two vertical forces and one horizontal.

the potentional energy= energy from the explosive.

No, that ignores energy lost to drag.

 

[vbottas]

I will attach an image. Give me just a sec

 

[vbottas]

 

[vbottas]

the front wheels are supposed to be touching the ground. makes 35 degree angle with the ground,.

 

[vbottas]

The only thing I need clarifying is the angle. Is that the launch angle, or the angle of the line from the launch point to its highest point?

No, the blast is a reminder one-off impulse. Once off the ground, there are just two vertical forces and one horizontal.

No, that ignores energy lost to drag.

Thank you so much for your help so far. But unfortunally, it is 2am where I am. So I am gonna have to call it a night. Hopefully we can pick it back up tomorrow!

 

[haruspex]

the front wheels are supposoed to be touching the ground. makes 35 degree angle with the ground,.

Glad you posted that… I assumed it was airborne.
So there continues to be a normal force from the ground (but no tractive force), and the drag force varies along the length of the roof. This is really messy. Approximations will be needed.

Not sure if you realise this, but if the x and y velocities are $\dot x, \dot y$ then the drag is proportional to something like $\dot x^2+\dot y^2$ (not exactly that because the coefficients are different), so its x and y components will be like $-\dot x\sqrt{\dot x^2+\dot y^2}, -\dot y\sqrt{\dot x^2+\dot y^2}$.

Note that if all its horizontal velocity were converted into GPE it would reach about 20m without drag. This implies the vertical velocity just after the explosion, even at the back, is much less than the horizontal velocity. So you can make an approximation for vertical drag on the basis that $\dot y<<\dot x$.