Question

The Bellagio Hotel in Las Vegas, Nevada, is well known for its Musical Fountains, which use 192 HyperShooters to fire water hundreds of feet into the air to the rhythm of music. One of the HyperShooters fires water straight upward to a height of \(240 \mathrm{ft}\). a) What is the initial speed of the water? b) What is the speed of the water when it is at half this height on its way down? c) How long will it take for the water to fall back to its original height from half its maximum height?

Short answer

Answer: The initial speed of the water shot by the HyperShooter is 80 ft/s, the speed of the water at half-height on its way down is approximately 59.4 ft/s, and it takes approximately 3.885 seconds for the water to fall back to its original height from half its maximum height.

Step-by-step solution

Part a: Finding the initial speed

To find the initial speed of the water, we will use the first equation of motion: \(v^2 = u^2 + 2as\) We are given the maximum height, \(s = 240 \mathrm{ft}\), and we know that the final velocity at the top, \(v\), is \(0 \mathrm{ft/s}\). Plugging these values and acceleration \(a = -32 \mathrm{ft/s^2}\) into the equation, we get: \(0 = u^2 + 2 \times (-32) \times 240\) Solving for \(u\), we find: \(u^2 = 2 \times 32 \times 240\) \(u = \sqrt{2 \times 32 \times 240}\) \(u = 80 \mathrm{ft/s}\) So, the initial speed of the water is \(80 \mathrm{ft/s}\).

Part b: Finding the speed of the water at half-height on its way down

When the water is at half its maximum height on its way down, \(s = 120 \mathrm{ft}\). We can use the first equation of motion again, with \(s=120\) and \(v\) being the velocity at half-height: \(v^2 = u^2 + 2as\) \((80)^2 - 2(-32)(120) = v^2\) \(v = \sqrt{(80)^2 - 2(-32)(120)}\) \(v \approx 59.4 \mathrm{ft/s}\) Thus, the speed of the water at half-height on its way down is approximately \(59.4 \mathrm{ft/s}\).

Part c: Finding the time for the water to fall back to its original height from half its maximum height

We are now asked to find the time it takes for the water to fall back to its original height from half its maximum height. We will use the second equation of motion: \(s = ut + \frac{1}{2}at^2\) At half-height on its way down, \(s = 120 \mathrm{ft}\) and \(u = 59.4 \mathrm{ft/s}\). Plugging these values, along with \(a = -32 \mathrm{ft/s^2}\) into the equation, we get: \(120 = 59.4t + \frac{1}{2}(-32)t^2\) To solve this quadratic equation for \(t\), we can use the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where \(a = -16\), \(b = 59.4\), and \(c = -120\). Plugging in the values, we obtain two solutions for \(t\): \(t_1 \approx 1.918\, \mathrm{s}\) \(t_2 \approx 3.885\, \mathrm{s}\) We are interested in the time it takes for the water to fall from half its maximum height to the ground (\(0 \mathrm{ft}\)). Since \(t_1\) corresponds to the time it takes for the water to reach half the height while going up, \(t_2\) represents the time it takes for the water to fall from half its maximum height to the ground. Therefore, it takes approximately \(3.885 \mathrm{s}\) for the water to fall back to its original height from half its maximum height.

Key concepts

Equations of Motion

The study of projectile motion in physics often involves understanding the equations of motion, which describe the path of an object under the influence of gravity. When an object like the water from the Bellagio Hotel HyperShooters is projected vertically, its motion can be predicted and analyzed using these crucial equations.

First, let's recall the basic equations of motion. The one most relevant to our HyperShooter problem is:
\(v^2 = u^2 + 2as\), where \(v\) represents the final velocity, \(u\) is the initial velocity, \(s\) is the displacement, and \(a\) is the acceleration due to gravity. By applying this equation, we can calculate various aspects of the water's projectile motion, such as the initial speed required to reach a certain height.

In the context of the HyperShooter, using this first equation allowed us to determine that the water reaches the peak of its trajectory (\