Question

A stone is released from the top of the tower of height 19.6m . Calculate its final velocity just before it touches the ground.​

Answer 1

Given :-

• Height of tower (s) = 19.6 m

• Initial velocity (u) = 0 m/s

• Acceleration due to gravity (g) = 9.8 m/s²

To find :-

• Final velocity (v) = ?

Solution :-

we know that,

$\blue{\bigstar}\:\:{\underline{\boxed{\bf\red{{v}^{2}={u}^{2}+2gs}}}}$

$\rm\longrightarrow\:{v}^{2}={(0)}^{2}+2\times{9.8}\times{19.6}$

$\rm\longrightarrow\:{v}^{2}=19.6\times{19.6}$

$\rm\longrightarrow\:{v}^{2}={(19.6)}^{2}$

$\rm\longrightarrow\:v=19.6\:m/s$

Hence,the final velocity of the stone just before touching the stone will be 19.6 m/s.

More Information :-

Acceleration due to gravity :- The acceleration produced on a free falling body due to gravitational force is known as acceleration due to gravity.

• It is denoted by g
• It change from place to place.
• It's unit are m/s².

Answer 2

Given ,

Initial velocity (u) = 0 m/s

Displacement (s) = 19.6 m

Acceleration due to gravity (a) = 9.8 m/s²

We know that , the Newton's third equation of motion is given by

$\boxed{ \sf{(v)}^{2} - {(u)}^{2} = 2as }$

Thus ,

(v)² - (0)² = 2 × 9.8 × 19.6

(v)² = (19.6)²

Taking square root on both sides , we get

v = 19.6 m/s

Therefore ,

The final velocity just before it touches the ground is 19.6 m/s