Answer:
6.0 s
98 m/s
Explanation:
The radius of the planet is much bigger than the height of the tower, so we will assume the acceleration is constant. Neglect air resistance.
Acceleration due to gravity on this planet is:
a = GM / r²
a = (6.67×10⁻¹¹ m³/kg/s²) (2.7 × 1.48×10²³ kg) / (1.7 × 750,000 m)²
a = 16.4 m/s²
The height of the tower is:
Δy = 96 × 3.05 m
Δy = 293 m
Given v₀ = 0 m/s, find t and v.
Δy = v₀ t + ½ at²
(293 m) = (0 m/s) t + ½ (16.4 m/s²) t²
t = 6.0 s
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (16.4 m/s²) (293 m)
v = 98 m/s
