Question

A 16.0 kg rock is at the top of a 42.0 m. tall hill and is dropped off the cliff. At what speed will the rock hit the ground? (HINT: Think energy conservation)

Answer 1

Provided that:

• Mass = 16.0 kg
• Height = 42.0 metres

To calculate:

• The speed

Solution:

At the speed of 28.98 ≈ 29 metre per second the rock will hit the ground when it is dropped off the cliff from a tall hill

Using concept:

• Energy conservation

Using dimension:

• ${\small{\underline{\boxed{\sf{\dfrac{1}{2} \: mv^2 \: = mgh}}}}}$

Where, m denotes mass, v denotes velocity, g denotes acceleration due to the gravity and h denotes height

Knowledge required:

↪️ g denotes acceleration due to gravity.

↪️ The universal value of g is 9.8 metre per second sq. but let us take g as 10 metre per second sq. for easy calculation as we are not asked to take g as 9.8 metre per second sq.

Required solution:

$:\implies \sf \dfrac{1}{2} \: mv^2 \: = mgh \\ \\ :\implies \sf \dfrac{1}{2} \not{m} v^2 \: = \not{m} gh \\ \\ :\implies \sf \dfrac{1}{2} v^2 \: = gh \\ \\ :\implies \sf \dfrac{v^2}{2} \: = gh \\ \\ :\implies \sf \dfrac{v^2}{2} \: = 10(42) \\ \\ :\implies \sf \dfrac{v^2}{2} \: = 420 \\ \\ :\implies \sf v^2 \: = 420 \times 2 \\ \\ :\implies \sf v^2 \: = 840 \\ \\ :\implies \sf \sqrt{v} \: = 840 \\ \\ :\implies \sf v \: = 28.98 \: ms^{-1} \\ \\ :\implies \sf v \: \approx \: 29 \: ms^{-1} \\ \\ :\implies \sf Velocity \: = 29 \: ms^{-1}$