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. 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

_/\_Hello mate__here is your answer--

u = 0 m/s

v = ?

s = Height of the stone = 19.6 m

g = 9.8 ms−2

According to the equation of motion under gravity

v^2−u^2=2gs

v^2−(0)^2=2×9.8×19.6

v^2=2×9.8 ×19.6=(19.6)^2

v=19.6 ms−1

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

I hope, this will help you.☺

Thank you______❤

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Answer 2

➣ Diagram:

In the above attachment.

➣ Question:

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

➣ Concept:

Here, in this given query , Height (s) , Initial velocity, (u) , Acceleration, (a) is given and we have to find the final velocity (v) . Since the Stone is released from the top of a tower so the Initial velocity will be 0 m/s. As we know, Distance travelled ⇢ Height ( so (s) will be 19.6 m ) . The acceleration will be positive as the stone fall downwards. By applying the formula of 3rd equation of motion we will get the Final velocity (v).

➣ Given:

• ❂ Height, s ➝ 19.6 m

• ❂ Initial velocity, u ➝ 0 m/s

• ❂ Acceleration, a ➝ 9.8 m/s²

➣ To Find:

• ❂ Final velocity, v ➝ ?

➣ Formula Required:

• v² - u² ➛ 2as

( 3rd Equation of Motion )

➣ Substitution:

v² - u² ➛ 2as

⟹ v² - (0)² ➛ 2 × 9.8 × 19.8

⟹ v² ➛ 19.6 × 19.6

⟹ v² ➛ 384.16

⟹ v² ➛ √384.16

⟹ v² ➛ 19.6 m/s

Hence, the final velocity is 19.8 m/s

➣ Explore:

Here are the Other Formulae:-

First equation of motion:-

• v ➛ u + at

Second equation of motion:-

• s ➛ ut + 1/2 at²

Third equation of motion:-

• v² - u² ➛ 2as