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Chapter 2: Problem 52

please assume the free-fall acceleration \(g=9.80 \mathrm{m} / \mathrm{s}^{2}\) unless a more precise value is given in the problem statement. Ignore air resistance. A balloonist, riding in the basket of a hot air balloon that is rising vertically with a constant velocity of \(10.0 \mathrm{m} / \mathrm{s},\) releases a sandbag when the balloon is \(40.8 \mathrm{m}\) above the ground. What is the bag's speed when it hits the ground?

Short Answer

Expert verified
Answer: The speed of the sandbag when it hits the ground is approximately \(15.2\mathrm{m}/\mathrm{s}\) in the downward direction.

Step by step solution

01

Identify the given information

We are given the following information: 1. The initial height of the sandbag (\(h_0\)) is \(40.8 \mathrm{m}\) above the ground. 2. The initial vertical velocity of the sandbag (\(v_0\)) is the same as the hot air balloon, which is \(10.0 \mathrm{m} / \mathrm{s}\) (upward direction). 3. The acceleration due to gravity (\(g\)) is given as \(-9.80 \mathrm{m} / \mathrm{s}^{2}\) (since it's acting downwards).
02

Find the time it takes for the sandbag to hit the ground

We can use the following equation for a uniformly accelerated motion to find the time it takes for the sandbag to hit the ground: \(h(t) = h_0 + v_0t + \frac{1}{2}at^2\) Since the sandbag hits the ground, \(h(t) = 0\) and we can solve for \(t\): \(0 = 40.8 + 10t - 4.9t^2\) This is a quadratic equation in the form of \(at^2 + bt + c = 0\) with \(a = -4.9\), \(b = 10\), and \(c = 40.8\). We can use the quadratic formula to find the roots: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
03

Solve for the time when the sandbag hits the ground

Substitute the given values and solve for t: \(t = \frac{-10 \pm \sqrt{10^2 - 4(-4.9)(40.8)}}{2(-4.9)}\) \(t \approx 2.57\mathrm{s}\) (Only consider the positive root as time cannot be negative)
04

Calculate the final speed of the sandbag

Now that we have the time it takes for the sandbag to hit the ground, we can find the final speed of the sandbag using the equation for uniformly accelerated motion: \(v(t) = v_0 + at\) Plugging in the given values: \(v(2.57) = 10 - 9.80(2.57)\) \(v(2.57) \approx -15.2\mathrm{m}/\mathrm{s}\) (The negative sign indicates the downward direction) The speed of the sandbag when it hits the ground is approximately \(15.2\mathrm{m}/\mathrm{s}\) in the downward direction.

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