How Can I Calculate Deceleration and Depth for an Object Falling into Water?

  • Thread starter Thread starter lalapsambel
  • Start date Start date
  • Tags Tags
    Resistance Water
AI Thread Summary
To calculate the deceleration and depth of an object falling into water, the impulse equation F.t = m(Vt - Vo) is not suitable as it doesn't account for water resistance. Instead, use the drag force equation F = 1/2 * ρ * A * Cd * v^2 to determine the force of water resistance, where ρ is water density, A is cross-sectional area, Cd is the drag coefficient, and v is velocity. The deceleration can then be found using a = F/m, where F is the drag force and m is the object's mass. For depth, apply d = 1/2 * a * t^2, with a being the calculated deceleration and t the time to stop. Accurate calculations depend on proper use of these equations and careful attention to units.
lalapsambel
Messages
1
Reaction score
0
Here's my problem:
I have to find how much deceleration & how deep it will be when something fall into the water, but I just have the Vo when it strikes the water and of course Vt which is 0. I don't know how can I measure its deceleration underwater. OK, actually I just want to know how much force that the water gives when this thing strikes the water. So, I want to combine the equation F.t= m(Vt-Vo). Can I use that? But I have to get either the time and I'll use Vt=0, or the Vt and I'll use the time = 0.01 s, so it's just instanteous velocity.

What equation should I use? Please help me, it's a part of my project and if I can't solve this problem, it mean I should change the whole project!
 
Physics news on Phys.org
For any constant acceleration, you can treat "average" as (v0+v1)/2 - i.e. the average of the initial and final speeds. Here, v1= 0 so the average velocity is v0/2.

Decelerating at constant acceleration, -a, v(t)= v0- at so v(t)= 0 when v0- at= 0 or t= v0/a. The distance the object falls in that time is (v0/2)(v0/a)= v02/2a.

If you know the initial velocity v0 and depth d, then
d= v02/2a so a= v02/2d.
 


Hi there,

I understand your frustration with this water resistance problem. It can be challenging to calculate the deceleration and depth of an object falling into water without all the necessary information. However, there are a few things you can consider to help you solve this problem.

First, it's important to note that the equation you mentioned, F.t= m(Vt-Vo), is not the correct one to use in this situation. This equation is used to calculate the impulse, or change in momentum, of an object. It does not take into account the force of water resistance.

To calculate the force of water resistance, you can use the equation F = 1/2 * ρ * A * Cd * v^2, where ρ is the density of water, A is the cross-sectional area of the object, Cd is the drag coefficient, and v is the velocity of the object. This equation will give you the force of water resistance acting on the object at a particular velocity.

To find the deceleration of the object, you can use the equation a = F/m, where F is the force of water resistance and m is the mass of the object. This will give you the deceleration of the object as it falls into the water.

As for determining the depth of the object, you can use the equation d = 1/2 * a * t^2, where d is the depth, a is the deceleration calculated in the previous step, and t is the time it takes for the object to come to a complete stop (which can be estimated based on the initial velocity and the force of water resistance).

I hope this helps you with your problem. Remember to double-check your equations and units to ensure accuracy. Good luck with your project!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top