Questions: A student is skateboarding down a ramp that is 7.46 m long and inclined at 17.3° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 3.82 m / s. Neglect friction and find the speed at the bottom of the ramp.

Solution Steps

We need to find the speed of the skateboarder at the bottom of the ramp. The given data includes:

• Length of the ramp, \( L = 7.46 \, \text{m} \)
• Incline angle, \( \theta = 17.3^\circ \)
• Initial speed at the top, \( v_i = 3.82 \, \text{m/s} \)

Since friction is neglected, we can use the conservation of mechanical energy. The total mechanical energy at the top of the ramp is equal to the total mechanical energy at the bottom of the ramp.

The mechanical energy at the top consists of kinetic energy and potential energy: \[ E_{\text{top}} = \frac{1}{2} m v_i^2 + mgh \] where \( h \) is the height of the ramp.

At the bottom of the ramp, the potential energy is zero, and the energy is purely kinetic: \[ E_{\text{bottom}} = \frac{1}{2} m v_f^2 \]

The height \( h \) of the ramp can be calculated using trigonometry: \[ h = L \sin(\theta) = 7.46 \times \sin(17.3^\circ) \]

Set the total energy at the top equal to the total energy at the bottom: \[ \frac{1}{2} m v_i^2 + mgh = \frac{1}{2} m v_f^2 \]

Cancel the mass \( m \) from both sides and solve for \( v_f \): \[ \frac{1}{2} v_i^2 + gh = \frac{1}{2} v_f^2 \]

\[ v_f^2 = v_i^2 + 2gh \]

\[ v_f = \sqrt{v_i^2 + 2gh} \]

Substitute the values: \[ v_f = \sqrt{3.82^2 + 2 \times 9.81 \times (7.46 \times \sin(17.3^\circ))} \]

Calculate the numerical value: \[ v_f = \sqrt{3.82^2 + 2 \times 9.81 \times (7.46 \times 0.2970)} \]

\[ v_f = \sqrt{14.5924 + 2 \times 9.81 \times 2.2166} \]

\[ v_f = \sqrt{14.5924 + 43.4781} \]

\[ v_f = \sqrt{58.0705} \]

\[ v_f \approx 7.6191 \, \text{m/s} \]

Final Answer