Question

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

Answer 1

Answer:-

Given:-

Initial velocity of the stone = 0 m/s  

Final velocity of the stone

Height of the stone = 19.6 m

Acceleration due to gravity = 9.8 ms−2

Solution:-

To find this we use, the equation of motion under gravity v²−u² = 2gs  

∴ v²− 0²= 2 × 9.8 × 19.6  

⇒ v² = 2 × 9.8 × 19.6 = (19.6)2  

⇒ v² = √384.16

⇒ v = 19.6 ms−1  

Hence, the velocity of the stone just before touching the ground is 19.6 ms−1 .

Answer 2

Given:

Initial velocity of stone (u) = 0 m/S

Height of the tower (s) = 19.6 m

Acceleration due to gravity = 9.8 m/s²

To find:

We have to find the final velocity of stone just before touching the ground.

Solution:

To find the velocity, we will use third equation of motion:

v² - u² = 2gs

⇒ v² - 0² = 2 * 9.8 * 19.6

⇒ v² = 384.16

⇒ v = √384.16

⇒ v = 19.6 m

Hence, the final velocity of stone just before touching the ground- 19.6 m/s.