A Balloon carrying a basket is descending at a constant velocity of 20.0 m/s. A person in the basket throws a stone with an initial velocity of 15.0 m/s horizontally perpendicular to the path of descending balloon and 4.00 (s) later this person sees the stone strike the ground.

a) How high was the balloon when the stone was thrown out?
b) How far horizontally does the stone travel before it hits the ground?
c) At the instant the stone hits the ground, how far is it from the basket?

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(II) You are explaining why astronauts feel weightless while orbiting in the space shuttle. your friends respond that they thought gravity was just alot weaker up there. convince them and yourself that it isn't so by calculating the acceleration of gravity 250 km above the Earth's surface in term of g.

To solve this problem, we need to use the information given and apply some principles of physics such as kinematics. Let's break down each part of the problem:

a) How high was the balloon when the stone was thrown out?

To determine the height of the balloon when the stone was thrown out, we can use the equation of motion:

\[h = h_0 + v_0t + \frac{1}{2}gt^2\]

Where:
- \(h\) is the final height of the balloon
- \(h_0\) is the initial height of the balloon
- \(v_0\) is the initial velocity of the stone (horizontal velocity)
- \(t\) is the time elapsed (4.00 seconds)
- \(g\) is the acceleration due to gravity (-9.8 m/s^2)

Since the stone is thrown out horizontally, its initial vertical velocity is zero. Therefore, the equation simplifies to:

\[h = h_0 + \frac{1}{2}gt^2\]

Substituting the known values:

\[h = h_0 + \frac{1}{2}(-9.8)(4.00)^2\]

Solve for \(h_0\) to find the initial height of the balloon.

b) How far horizontally does the stone travel before it hits the ground?

Since the horizontal velocity of the stone remains constant at 15.0 m/s, and the time it takes for the stone to hit the ground is 4.00 seconds, we can find the horizontal distance traveled using the formula:

\[d = v \cdot t\]

Where:
- \(d\) is the horizontal distance traveled by the stone
- \(v\) is the horizontal velocity of the stone (15.0 m/s)
- \(t\) is the time elapsed (4.00 seconds)

Solve for \(d\) to find the horizontal distance traveled by the stone.

c) At the instant the stone hits the ground, how far is it from the basket?

Since the stone is thrown horizontally perpendicular to the path of the descending balloon, the stone will travel straight horizontally. Therefore, the horizontal distance traveled by the stone is the distance between the stone and the basket at the moment the stone hits the ground. This is the same distance we calculated in part (b).

Solve for \(d\) to find the horizontal distance between the stone and the basket.

By following these steps, you will be able to answer all three parts of the problem. Let me know if you need further help with the calculations.