A bobsled makes a run down an ice track starting at 150 m vertical distance up the hill. If there is no friction, what is the velocity at the bottom of the hill? 1) \( 27 \mathrm{~m} / \mathrm{s} \) 2) \( 36 \mathrm{~m} / \mathrm{s} \) 3) \( 45 \mathrm{~m} / \mathrm{s} \) 4) \( 54 \mathrm{~m} / \mathrm{s} \) 5) \( 65 \mathrm{~m} / \mathrm{s} \)

To find the velocity of the bobsled at the bottom of the hill, we can use the principle of conservation of energy. The potential energy (PE) at the top is converted into kinetic energy (KE) at the bottom. The potential energy can be calculated using the formula: \[ \text{PE} = mgh \] where \( m \) is mass, \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and \( h \) is the height (150 m in this case). At the bottom, all this energy is converted into kinetic energy, given by: \[ \text{KE} = \frac{1}{2} mv^2 \] Since mass \( m \) cancels out, we can set the two equations equal to each other: \[ mgh = \frac{1}{2} mv^2 \] Simplifying gives: \[ gh = \frac{1}{2} v^2 \] Solving for velocity \( v \): \[ v^2 = 2gh \\ v = \sqrt{2gh} \] Substituting in values: \[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 150 \, \text{m}} \\ v = \sqrt{2943} \\ v \approx 54.27 \, \text{m/s} \] Thus, the closest answer is: 4) \( 54 \mathrm{~m/s} \)