Crate Freefall Calculation: Air Drag at 2,000m Altitude

In summary, the Crate Freefall Calculation at 2,000m Altitude takes into account the effects of air drag on a falling crate. This calculation involves determining the velocity and acceleration of the crate as it falls, as well as the drag force acting upon it due to air resistance. By considering the crate's weight, surface area, and air density at the given altitude, the calculation can estimate the time it will take for the crate to reach the ground. This information is crucial for ensuring the safety and success of aerial deliveries and drops.
  • #1
espo
a crate falls from an airplane flying horizontally at an altitude of 2,000 m. neglecting air drag, howlong will the crate take to strike the ground/ is the formula t = d/v help
 
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  • #2
Code:
First write out what you know:

u = 0 m/s
v = v
a = 9.81 m/s[sup]2[/sup]
s = 2000 m
t = t

then pick the appropriate formula:

s = ut + 0.5at[sup]2[/sup]
2000 = 4.905 * t[sup]2[/sup]
t[sup]2[/sup] = 407.7
t = 20.2 s
 
  • #3


Neglecting air drag, the crate will take approximately 14.14 seconds to strike the ground. This calculation can be determined using the formula t = d/v, where t is the time, d is the distance, and v is the velocity. In this scenario, the distance is 2,000 m (the altitude) and the velocity is the acceleration due to gravity, which is approximately 9.8 m/s^2. Therefore, t = 2,000/9.8 = 14.14 seconds. This formula is helpful in determining the time it takes for an object to fall a certain distance under the influence of gravity, but it does not account for air resistance or drag. In reality, the crate would experience air resistance, which would slow down its descent and affect the actual time it takes to strike the ground. To calculate the exact time, the air resistance would need to be taken into consideration.
 

1. How is air drag calculated during a crate freefall at 2,000m altitude?

The air drag during a crate freefall at 2,000m altitude is calculated using the equation: drag force = 0.5 * air density * velocity^2 * drag coefficient * frontal area. The air density and drag coefficient are dependent on the air temperature and the shape of the crate, respectively. The velocity can be calculated using the equation: velocity = square root of (2 * acceleration due to gravity * altitude).

2. What is the significance of air drag in a crate freefall at 2,000m altitude?

Air drag plays a crucial role in the descent of a crate in freefall at 2,000m altitude. It is the force that opposes the motion of the crate, slowing it down and causing it to reach terminal velocity. This allows for a controlled descent and prevents the crate from reaching dangerously high speeds.

3. How does air density affect the calculation of air drag at 2,000m altitude?

Air density is a key factor in calculating air drag during a crate freefall at 2,000m altitude. As altitude increases, the air density decreases, resulting in a decrease in air drag. This is because there are fewer air molecules to collide with the crate and create resistance. Therefore, the air density must be taken into account when calculating the drag force on the crate.

4. How does the shape of the crate affect the calculation of air drag at 2,000m altitude?

The shape of the crate has a significant impact on the calculation of air drag during a freefall at 2,000m altitude. The drag coefficient, which is a measure of how streamlined an object is, is heavily dependent on the shape of the crate. A more streamlined shape will have a lower drag coefficient and experience less air resistance, resulting in a lower drag force.

5. What other factors should be considered when calculating air drag during a crate freefall at 2,000m altitude?

In addition to air density and crate shape, other factors that should be considered when calculating air drag during a crate freefall at 2,000m altitude include wind speed and direction, air temperature, and altitude change over time. These factors can all affect the drag force acting on the crate and should be taken into account for a more accurate calculation.

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