Calculating Drop Time from 105m in a Helicopter

In summary, the conversation is about calculating the time it takes for a box dropped from a helicopter to reach the ground. The problem is determining which equation to use and the initial height and velocity of the package. The equation used is xf=xi+vit+.5at^2, where xi is the initial height, vi is the initial velocity, and a is the acceleration due to gravity. It is confirmed that the helicopter's speed, 5.50m/s, is also the speed of the package. The initial height and velocity of the package are xi=0m, xf=105m, vi=5.50m/s, vf=0m/s, and a=9.80m/s^2.
  • #1
marsh
7
0
I need some help with a question please.
If a helicopter has a speed of 5.50m/s moving upward vertically and when at the height of 105m above Earth a box is dropped from the window how much time will it take to reach the ground?
 
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  • #2
You need to show some work first, and let us know where you are getting stuck.
 
  • #3
the problem is that i don't know what equation to use for this problem. Would the speed of the helicopter, 5.50m/s, be the speed of the package as well?
 
  • #4
do i use xf=xi+vit+.5at^2 ?
 
  • #5
x= xi+ vit+ (1/2)at2

where xi is the initial height, vi is the initial velocity and a is the acceleration due to gravity. Do you know what the initial height and velocity are for the package?

My answer to your question, which you posted while I was typing, is "yes"!
 
  • #6
so the xi= om xf=105m and vi=5.50m/s vf= 0m/s a= 9.80m/s^2
 
  • #7
is that correct?

or is the xf= 0m and xi=105m b/c it's being dropped from the top?
 
Last edited:

1. How do you calculate the drop time from 105m in a helicopter?

In order to calculate the drop time, you will need to use the equation t = √(2h/g), where t is the drop time, h is the height (105m in this case), and g is the acceleration due to gravity (9.8 m/s²). This equation assumes no air resistance and a constant acceleration due to gravity.

2. What is the value of g used in the calculation?

The value of g used in the calculation is the standard acceleration due to gravity on Earth, which is 9.8 m/s². This value may vary slightly depending on location and altitude, but for most practical purposes, 9.8 m/s² is a good approximation.

3. Can this equation be used for any height and any type of object?

Yes, this equation can be used for any height as long as no air resistance and a constant acceleration due to gravity are assumed. However, it may not be accurate for objects with a significantly different shape or mass compared to the typical human body.

4. Does the weight of the object affect the drop time?

No, the weight of the object does not affect the drop time. This equation assumes that all objects, regardless of weight, fall with the same acceleration due to gravity.

5. How accurate is this calculation in real-life scenarios?

This calculation is a theoretical approximation and may not be accurate in real-life scenarios due to factors such as air resistance and the varying acceleration due to gravity at different altitudes. It is important to note that this calculation is a simplified model and should be used as a guide, not an exact measurement.

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