一物體以9.8米/秒的初速從地面沿仰角30度的光滑斜面上運動

1. Use Law of Conservation of Mechanical Enery(機械能守恆定律)
kinetic energy (動能) at bottom of slope = potential energy (勢能) at highest point
hence, (1/2)m(9.8)^2 = mg.h
where m is mass of the object, g is the acceleration due to gravity (= 9.8 m/s^2)
h = 9.8^2/2g m = 9.8/2 m = 4.9 m

2. Frictional force = [(1/3).square-root(3)] x mg.cos(30) = mg/2
Retarding force acting on the object
= mg.sin(30) + mg/2
Decceleration = mg(sin(30) + 1/2)/m = g(sin(30) + 1/2) = g

Use equation of motion: v^2 = u^2 + 2a.s
with v = 0 m/s, u = 9.8 m/s, a = -g(=-9.8 m/s^2), s =?
hence, 0 = 9.8^2 + 2(-9.8)s
s = 4.9 m

Vertical height reached = 4.9.sin(30) m = 2.45 m