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.


・Class 9th - Chapter Gravitation​

Answer 1

Answer:

your answer : 19.6m/s

Explanation:

Given :

u = 0m/s [taken]

h = 19.6m

g = 9.6m/s [taken]

v = ?

to find final velocity , we apply 3rd equation of law of motion :-

[ acceleration due to gravity ]

v^2 = u^2 + 2gh

v^2 = 0 + 2*9.6m/s * 19.6m

v^2 = √ 19.6 * 19.6 m/s

v = 19.6m/s

hence , the final velocity before touching the ground is 19.6m/s .

Answer 2

Answer :-

Given -

• Intial Velocity (u) = 0m/s
• Height of the stone (s) = 19.6m
• Acceleration due to gravity (g) = 9.8 ms^-2

We have to find-

• Final velocity (v)

_____....

According to the equation of motion under gravity -

v² − u² =2gs

∴ v² − 0² = 2 × 9.8 × 19.6

⇒ v² = 2 × 9.8 × 19.6

⇒ v² = (19.6)² [ since, 9.8×2= 19.6]

⇒ v = 19.6 ms^-1

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