No. A neutron star is the weird remains of a star that blew its outer layers off
in a nova event, and then had enough mass left so that gravity crushed its
electrons into its protons, and then what was left of it shrank down to a sphere
of unimaginably dense neutron soup. But it didn't have enough mass to go
any farther than that.
A black hole is the remains of a star that had enough mass to go even farther
than that. No force in the universe was able to stop it from contracting, so it
kept contracting until its mass occupied no volume ... zero. It became even
more weird, and is composed of a substance that we don't know anything about
and can't describe, and occupies zero volume.
Contrary to popular fairy tales, a black hole doesn't reach out and "suck things in".
It's just so small (zero) that things can get very close to it. You know that gravity
gets stronger as you get closer to an object, so if the object has no size at all, you
can get really really close to it, and THAT's where the gravity gets really strong.
You may weigh, let's say, 100 pounds on the Earth. But you're like 4,000 miles
from the center of the Earth. What if all of the earth's mass was crammed into
the size of a bean. Then you could get 1 inch from it, and at that distance from
the mass of the Earth, you would weigh 25,344,000,000 pounds.
But Earth's mass is not enough to make a black hole. That takes a minimum
of about 3 times the mass of the sun, which is right about 1 million times the
Earth's mass. THEN you can get a lightweight black hole.
Do you see how it works now ?
I know. It all seems too fantastic to be true.
It sure does.
Answer:
It would take the object 5.4 s to reach the ground.
Explanation:
Hi there!
The equation of the height of a free-falling object at any given time, neglecting air resistance, is the following:
h = h0 + v0 · t + 1/2 · g · t²
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
g = acceleration due to gravity (-32.2 ft/s² considering the upward direction as positive).
t = time
Let´s supose that the object is dropped and not thrown so that v0 = 0. Then:
h = h0 + 1/2 · g · t²
We have to find the time at which h = 0:
0 = 470 ft - 1/2 · 32.2 ft/s² · t²
Solving for t:
-470 ft = -16.1 ft/s² · t²
-470 ft / -16.1 ft/s² = t²
t = 5.4 s
