That will depend on lots of things; mainly, from how high it falls, what's its initial velocity, as well as its shape and size (which will affect air resistance).
The answer will depend on what "it" is, and on what its initial velocity is.
2s
if you ignore air resistance, it would take about 3.5 seconds; at ttat point where it hits the ground it is traveling at 75 mph. Because of the air drag, it may take a bit longer to hit the ground.
This is a velocity question so u need to use uvaxt
Assuming they were in a vacuum, if both objects were dropped from th esame height, then both take the same length of time to reach the ground. All masses fall with the same acceleration, reach the same speed in the same period of time, and hit the ground at the same time. Otherwise and if there is an atmosphere or if they are dropped from different heights, you have not presented information; shape and size are the most important factors.
A long time...
it takes 7 minutes
5 seconds
The answer will depend on what "it" is, and on what its initial velocity is.
Once it starts to develop a tornado may take any where from several minutes to a few seconds to reach the ground.
8.1 seconds
it will take about 30 seconds to 1 whole minute.
The longer you delay, the faster your speed when you hit the ground.
2s
The ball would take approximately 4 seconds to hit the ground, neglecting air resistance. This calculation is based on the constant acceleration due to gravity (9.8 m/s²), assuming the ball is dropped and not thrown.
if you ignore air resistance, it would take about 3.5 seconds; at ttat point where it hits the ground it is traveling at 75 mph. Because of the air drag, it may take a bit longer to hit the ground.
Fireworks are designed to explode in the sky, producing dazzling colors and shapes. If a firework were to somehow reach a star, it would be burned up long before getting close due to the intense heat and radiation emitted by stars. Stars are incredibly hot and massive celestial bodies, much larger and more powerful than fireworks.