If you did this then it could lead to cheating or someone else getting hurt.
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft
Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision
hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)