Question

A stone is released from the top of a tower of height 50 m. Calculate its final velocity just
before touching the ground.
​

Answer 1

Given

A stone is released from the top of a tower of height 50 m

To Find

Its final velocity just  before touching the ground

Solution

We know that

Equation of motion (under gravity)

$\red{\sf \longrightarrow v^{2}-u^{2}=2gs}$

Here

• v = final velocity
• u = initial velocity
• g = acceleration due to gravity
• s = height

Units

• v = metres per second (m/s)
• u = metres per second (m/s)
• g = metres per second square (m/s²)
• s = metre (m)

According to the question

We are asked to find its final velocity just  before touching the ground

Therefore

We must find "v"

Given that

A stone is released from the top of a tower of height 50 m

Hence

• u = 0 m/s [starts from rest]
• s = 50 m

We know that

• g = 10 m/s²

Substituting the values

We get

$\sf \longrightarrow v^{2}-0^{2}=2 \times 10 \times 50$

$\sf \longrightarrow v^{2}=2 \times 10 \times 50$

$\sf \longrightarrow v^{2}=20 \times 50$

$\sf \longrightarrow v^{2}=1000$

$\sf \longrightarrow v=\sqrt{1000} \ m/s$

Therefore

$\pink{\sf \longrightarrow v=31.62 \ m/s}$

Hence

Final velocity just before touching the ground = 31.62 m/s