Studying Microgravity. NASA'S Glen Research Center in Cleveland, Ohio, has a 435-foot drop tower that begins on the surface and descends into Earth like a mineshaft.
How long will it take a sealed container to fall 435 feet? Round to the nearest tenth of a second.
the distance s = 1/2 at^2
You have s and a, so just find t.
I am sorry I don't understand? Is = 1/2 at^2 the answer? Thank you for you time and efforts Steve,
To determine the time it takes for a sealed container to fall 435 feet in NASA's drop tower, we can use the equation for free fall motion:
d = (1/2) * g * t^2
Where:
- d is the distance (435 feet in this case)
- g is the acceleration due to gravity (approximately 32.2 feet/second^2)
- t is the time we want to find
Rearranging the equation to solve for t, we have:
t^2 = (2 * d) / g
Substituting the given values, we get:
t^2 = (2 * 435) / 32.2
t^2 ≈ 27.018633540372674
Taking the square root of both sides to solve for t, we find:
t ≈ √(27.018633540372674)
t ≈ 5.2 seconds (rounded to the nearest tenth of a second)
Therefore, it will take approximately 5.2 seconds for the sealed container to fall 435 feet in NASA's drop tower.