The figure below shows a motorcycle leaving the end of a ramp with a speed of 37.9 m/s and following the curved path shown. At the peak of the path, a maximum height h above the top of the ramp, the motorcycle's speed is 36.0 m/s. What is the maximum height h? Ignore friction and air resistance. (Enter your answer in m.)

1 answer

To find the maximum height \( h \), we can use the principle of conservation of mechanical energy, which states that the total mechanical energy (kinetic energy + potential energy) remains constant if we ignore air resistance and friction.

  1. Initial Energy at the Ramp: When the motorcycle leaves the ramp:

    • Kinetic energy (\( KE_1 \)): \[ KE_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (37.9 , \text{m/s})^2 \]
    • Potential energy (\( PE_1 \)): Assuming the height at the ramp is \( 0 \): \[ PE_1 = 0 \]
    • Total energy at the ramp (\( E_1 \)): \[ E_1 = KE_1 + PE_1 = \frac{1}{2} m (37.9 , \text{m/s})^2 + 0 \]
  2. Energy at Maximum Height: At maximum height \( h \):

    • Kinetic energy (\( KE_2 \)): \[ KE_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (36.0 , \text{m/s})^2 \]
    • Potential energy (\( PE_2 \)): \[ PE_2 = mgh \]
    • Total energy at maximum height (\( E_2 \)): \[ E_2 = KE_2 + PE_2 = \frac{1}{2} m (36.0 , \text{m/s})^2 + mgh \]
  3. Setting the Energies Equal: Since total energy is conserved, we set \( E_1 = E_2 \): \[ \frac{1}{2} m (37.9 , \text{m/s})^2 = \frac{1}{2} m (36.0 , \text{m/s})^2 + mgh \]

  4. Cancelling Mass \( m \) (it appears in every term): \[ \frac{1}{2} (37.9 , \text{m/s})^2 = \frac{1}{2} (36.0 , \text{m/s})^2 + gh \]

  5. Rearranging for \( h \): \[ gh = \frac{1}{2} (37.9 , \text{m/s})^2 - \frac{1}{2} (36.0 , \text{m/s})^2 \]

  6. Calculating \( g \) (acceleration due to gravity): \[ g = 9.81 , \text{m/s}^2 \]

  7. Substituting the values: \[ gh = \frac{1}{2} \left( (37.9)^2 - (36.0)^2 \right) \] \[ (37.9)^2 = 1435.41 \] \[ (36.0)^2 = 1296 \] \[ gh = \frac{1}{2} (1435.41 - 1296) \] \[ gh = \frac{1}{2} (139.41) = 69.705 , \text{J} \]

  8. Solving for \( h \): \[ h = \frac{69.705}{9.81} \] \[ h \approx 7.10 , \text{m} \]

Therefore, the maximum height \( h \) is approximately 7.10 meters.