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A 124 kg balloon carrying a 22 kg basket is descending with a constant downward velocity of 20 m/s. A 1 kg stone is thrown from the basket with an initial velocity of 15 m/s perpendicular to the path of descent. A person below measures that it takes the stone 5 s to hit the ground. At the instant the stone hits the ground, how far is it from the balloon?

Answer :

Final answer:

The stone is 75 meters away from the balloon when it hits the ground.

Explanation:

To solve this problem, we can start by calculating the acceleration of the stone using the formula: acceleration = (final velocity - initial velocity) / time. Since the stone is thrown horizontally, its initial velocity and final velocity in the vertical direction will be 0 m/s. Thus, the acceleration of the stone is 9.8 m/s2 (due to gravity).

Next, we can calculate the time it takes for the stone to hit the ground using the equation: distance = initial velocity × time + (0.5 × acceleration × time2). Plugging in the values, we get: distance = 15 m/s × 5 s + (0.5 × 9.8 m/s2 × 5 s2) = 75 m.

Since the stone was thrown horizontally, it maintains its horizontal velocity of 15 m/s. Therefore, the horizontal distance the stone travels from the balloon can be calculated using the formula: distance = velocity × time = 15 m/s × 5 s = 75 m. Therefore, the stone is 75 meters away from the balloon when it hits the ground.

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