Calculating Maximum Height of Object Thrown Upward

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In summary, the conversation discusses an equation for the movement of a mass thrown upward with initial velocity Z, taking into account gravity and air resistance. The equation is m(dv/dt) = (Z-g)m - Av. To calculate the maximum height, one can integrate v(t) over time. However, there is some confusion about the equation due to the (Z-g) term and the definition of air resistance. A suggested equation for this scenario is s = ut + 1/2 gt^2.
  • #1
Yann
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I don't know much about classical physics, if I throw a rock or mass M upward with an initial velocity Z, with gravity g and air resistance A acting against it, is this equation the right one ;

\(\displaystyle m\frac{dv}{dt} = (Z-g)m -Av\)

Then if I want to know the maximal height, I just need to integrate v(t) to find the distance at time t. Right ?

I'm really not sure about this equation, mostly because of the (Z-g) term. After all, it's about INITIAL velocity.
 
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  • #2
You have to define what you mean by air resistance 'A', is this a force or a drag coefficent, is it a function of V or a constant?
Generally the equation you want is s = ut + 1/2 gt^2

See the formulae sticky thread for more equations.
 
  • #3
If you define air resistance to be a force proportional to the velocity

Then the forces on it as it's going up are mg down and Av down, so you get

[tex]\frac{dv}{dt} = -g - \frac{Av}{m}[/tex]

You then have to solve this differential equation for v(t), using v(0) = Z as a boundary condition. Then you have all the information you need about the rock's movement.
 

1. How do you calculate the maximum height of an object thrown upward?

The maximum height of an object thrown upward can be calculated using the formula:
h = (v2sin2θ) / 2g
Where h is the maximum height, v is the initial velocity of the object, θ is the angle at which the object is thrown, and g is the acceleration due to gravity (9.8 m/s2).

2. What is the initial velocity in the formula for calculating maximum height?

The initial velocity (v) in the formula represents the initial speed at which the object is thrown upward. It can be measured in meters per second (m/s) or feet per second (ft/s).

3. How does the angle at which the object is thrown affect its maximum height?

The angle (θ) at which the object is thrown affects its maximum height by determining the vertical component of its initial velocity. The higher the angle, the greater the vertical component and therefore, the higher the maximum height. An angle of 90 degrees (straight up) will result in the maximum height of the object.

4. Is the formula for calculating maximum height only applicable to objects thrown upward?

Yes, the formula for calculating maximum height is specifically for objects thrown upward against the force of gravity. It does not account for other factors such as air resistance.

5. Can the formula for calculating maximum height be used for objects thrown on an incline?

No, the formula for calculating maximum height is only applicable for objects thrown vertically. For objects thrown on an incline, the maximum height can be calculated using the formula:
h = (v2sin2θ) / 2(g + asinθ)
Where a is the acceleration due to the incline and θ is the angle of the incline.

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