Question

A 5.0kg rock is dropped off a 15 m cliff. what will be the velocity as it hits the ground?. ​

Answer 1

Apply law of conservation of energy

• PE=KE
• mgh=1/2mv²
• 2gh=v²
• v²=2(10)(15)
• v²=300
• v=17.3m/s

Answer 2

17.3 m/s (1 d.p.)

Step-by-step explanation:

Kinetic Energy

`E_k=\sf (1)/(2)mv^2`

where:

• 
  `E_k` = kinetic energy in joules (J)
• m = mass in kilograms (kg)
• v = speed in meters per second (m/s)

Gravitational potential energy

`E_p=\sf mgh`

where:

• 
  `E_p` = gravitational potential energy in joules (J)
• m = mass in kilograms (kg)
• g = gravitational field strength in newtons per kilogram (N/kg)
• h = change in height in meters (m)

Principle of Conservation of Energy

Gravitational potential energy at the top = kinetic energy at the bottom

`\implies \large \text{$ E_(p \sf (top))=E_(k \sf(bottom)) $}`

Given:

• m = 5.0 kg
• h = 15 m

The gravitational field strength of the Earth is 10 N/kg (10 newtons per kilogram), therefore:

• g = 10 N/kg

Substituting the values into the formula and solving for v:

`\implies \large \text{$ E_(p \sf (top))=E_(k \sf(bottom)) $}`

`\implies \sf mgh=(1)/(2)mv^2`

`\implies \sf (5.0)(10)(15)=(1)/(2)(5.0)v^2`

`\implies \sf 750=(5)/(2)v^2`

`\implies \sf v^2=300`

`\implies \sf v=√(300)`

`\implies \sf v=17.3\:\: m/s \:\: (1\:d.p.)`