1. Identify the given values:
- Initial height at point A, \( h_A = 140 \, \text{m} \)
- Height at point B, \( h_B = 95 \, \text{m} \)
- Height at point D, \( h_D = 80 \, \text{m} \)
- Initial velocity at point A, \( v_A = 0 \, \text{m/s} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
- Horizontal distance from B to C is 120 m, but not needed for energy calculations.
2. Apply the conservation of mechanical energy between points A and B:
\[
KE_A + PE_A = KE_B + PE_B
\]
Since \( KE_A = \frac{1}{2}mv_A^2 = 0 \) (initial velocity is 0),
\[
mgh_A = \frac{1}{2}mv_B^2 + mgh_B
\]
3. Cancel mass \( m \) from both sides:
\[
gh_A = \frac{1}{2}v_B^2 + gh_B
\]
4. Rearrange to solve for \( v_B^2 \):
\[
\frac{1}{2}v_B^2 = g(h_A - h_B)
\]
\[
v_B^2 = 2g(h_A - h_B)
\]
5. Substitute values:
\[
v_B^2 = 2 \times 9.8 \, \text{m/s}^2 \times (140 \, \text{m} - 95 \, \text{m}) = 2 \times 9.8 \times 45 = 882 \, \text{m}^2/\text{s}^2
\]
6. Calculate \( v_B \):
\[
v_B = \sqrt{882} \approx 29.7 \, \text{m/s}
\]
7. The given solution in the image uses \( 2 \times (440) = 880 \), which is close to 882, and approximates \( v_B \approx 30 \, \text{m/s} \). This is a rounding difference due to using \( 440 \) instead of \( 45 \times 9.8 = 441 \).
8. Final answer:
\[
v_B \approx 30 \, \text{m/s}
\]
Parent Tip: Review the logic above to help your child master the concept of roller coaster physics worksheet.
