Webassign - Dropping a Package From a Helicopter

  • Thread starter Thread starter berenmacbowma
  • Start date Start date
  • Tags Tags
    Helicopter
Click For Summary
SUMMARY

The discussion focuses on calculating the time it takes for a package dropped from a helicopter ascending at 2.00 m/s from a height of 160 m to reach the ground. The initial attempts incorrectly included the time for the helicopter's ascent, leading to erroneous total time calculations. The correct approach involves determining the additional height gained by the package after being released and then calculating the time to fall using kinematic equations, specifically accounting for the acceleration due to gravity at 9.8 m/s². The final solution requires summing the time for both the upward motion and the subsequent fall.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of uniformly accelerated motion
  • Familiarity with the concept of gravitational acceleration (9.8 m/s²)
  • Ability to perform basic algebraic manipulations
NEXT STEPS
  • Review kinematic equations for uniformly accelerated motion
  • Learn how to calculate the time of flight for objects under gravity
  • Explore the effects of initial velocity on projectile motion
  • Practice similar physics problems involving vertical motion and acceleration
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in solving problems related to projectile motion and gravitational effects.

berenmacbowma
Messages
7
Reaction score
0
Q: A helicopter is ascending vertically with a speed of 2.00 m/s. At a height of 160 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

Attempt:
[how long it takes to reach the height]
v=delta d/time
2=160/time
2(time)=160
time=80 seconds

[how long it takes to fall]
v=delta d/time
9.8=160/time
9.8(time)=160
time=16.326

80+16.326

96.3 s (which is wrong.)
 
Physics news on Phys.org
berenmacbowma said:
Attempt:
[how long it takes to reach the height]
v=delta d/time
2=160/time
2(time)=160
time=80 seconds
The time it takes for the helicopter to reach the given height is irrelevant. They just want the time it takes for the package to fall to the ground, starting from the moment it is dropped.

[how long it takes to fall]
v=delta d/time
9.8=160/time
9.8(time)=160
time=16.326
9.8 m/s^2 is the acceleration of the falling package, not the speed.

How can you express the position of the package as a function of time? It is uniformly accelerated motion. (You'll need to review your kinematic equations. Here's a list: Basic Equations of 1-D Kinematics)
 
berenmacbowma said:
Q: A helicopter is ascending vertically with a speed of 2.00 m/s. At a height of 160 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

Attempt:
[how long it takes to reach the height]
v=delta d/time
2=160/time
2(time)=160
time=80 seconds

Read the question again. When the package is released at 160m it is traveling upwards at same speed as the helicopter. It will carry on upwards for a bit. You need to work out how much higher it goes and add that to the 160m. Also calculate the time that phase takes.

Hint: Use one of the equations of motion to account the deceleration due to gravity.

[how long it takes to fall]
v=delta d/time
9.8=160/time
9.8(time)=160
time=16.326

80+16.326

96.3 s (which is wrong.)

Having worked out the height it reaches in part one you can then work out how long it takes to fall.

Then add the time for each phase together.

Hint: use one of the equations of motion to take into account the acceleration due to gravity.
 
the upward motion the package will have will be rather insignificant because the 2m/s speed isn't fast enough to make a difference in the problem, keep in mind this is a basic physics homework, the person that made the exercise wasn't probably thinking of the upward motion, since this is probably a high school exercise.
 
The difference is about 7%.
 
Nogueira said:
keep in mind this is a basic physics homework, the person that made the exercise wasn't probably thinking of the upward motion, since this is probably a high school exercise.
I suspect that the author of the problem was well aware of the upward motion--that's part of the point of the exercise. Such problems are standard fare.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
13
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K