Classical Mechanics - Curved Trajectory

The curvature of Mars is such that its surface drops a vertical distance of 2.0 meters for every 3600 meters tangent to the surface. In addition, the gravitational acceleration near its surface is 0.4 times that near the surface of Earth. What is the speed a golf ball would need to orbit Mars near the surface, ignoring the effects of air resistance?

A. 0.9 km/s
B. 1.8 km/s
C. 3.6 km/s
D. 4.5 km/s
E. 5.4 km/s

(GR9677 #22)

Solution:

y = ½ g_marst²

v_x = x/t 

Given:
y = 2
g_mars = 0.4 g 
x = 3600

2 = ½(0.4)(10)(t)²
t = 1 s
v_x = 3600/1 = 3600 m/s = 3.6 km/s

Answer: C