A 30lb sand bag falls off a hot air balloon floating stationary 5000ft above the ground what is the speed of the bag at the moment it hits the ground?
vf^2=2gh
calculte Vf
convert first 5000ft to meters.
To find the speed of the bag at the moment it hits the ground, we can use the principles of physics.
First, let's break down the problem:
Given:
- The weight of the bag is 30lb.
- The hot air balloon is floating stationary at a height of 5000ft above the ground.
We can assume that the only force acting on the bag is gravity. Therefore, we can apply the kinematic equations of motion to solve for the speed of the bag just before it hits the ground.
The kinematic equation we will use is:
v^2 = u^2 + 2as
where:
- v is the final velocity of the bag (the speed at which it hits the ground).
- u is the initial velocity (the speed of the bag when it is released from the hot air balloon). Since the bag is initially stationary, u = 0.
- a is the acceleration due to gravity, which is approximately 32.174 ft/s^2 (acceleration due to gravity varies slightly with altitude and location).
- s is the distance fallen by the bag, which is the height of the hot air balloon, 5000ft.
Plugging the given values into the equation:
v^2 = 0^2 + 2 * 32.174 ft/s^2 * 5000 ft
Simplifying:
v^2 ≈ 321740 ft^2/s^2
To find the value of v, we can take the square root of both sides:
v ≈ √(321740 ft^2/s^2)
Calculating the square root:
v ≈ 567.5 ft/s (rounded to one decimal place)
Therefore, the speed of the bag at the moment it hits the ground is approximately 567.5 ft/s.