(O Level Phy) A free fall question and its various ways of solving

Many problems have more than one way of solving. Here is one problem on a free falling object. Try to solve it yourself before looking at the three methods that follow.

Question: An apple at 4.05 m from the ground falls from rest. What is the speed at which it reaches the ground? Take g = 10 m/s²

[Solution]

Method 1: Using energy

Gravitational potential energy at the highest point = kinetic energy just before reaching the ground (Conservation of energy)
mgh = ½ mv²
gh = ½ v²
10 m/s² × 4.05 m = ½ v²
v² = 2 × 10 m/s² × 4.05 m 
 v = √(2 × 10 m/s² × 4.05 m) 
v = 9 m/s

Method 2: Drawing the speed-time graph

Gradient of the graph = 10 m/s²

(v – u)/t = 10 m/s²

(v – 0)/t = 10

v = 10t

t = v/10

Area under the shaded graph = 4.05 m

½ × v × t = 4.05

Substituting t = v/10,

½ × v × v/10 = 4.05

v² = 2 × 10 × 4.05

v = 9 m/s

Method 3: Using a suitable formula

v² = u² + 2as

where v = final velocity = ? m/s,

u = initial velocity = 0 since the apple is initially at rest,

a = g = +10 m/s² taking the downward direction to be positive,

s = distance covered = 4.05 m

v² = 0² + 2 × 10 m/s² × 4.05 m

v = 9 m/s


(Chinese translations: Conservation of energy 能量守恒, Gradient 坡度, Gravitational potential energy 重力势能, Kinetic energy 动能, velocity 速度)