Question

a stone is dropped from the top of a building.it's acceleration is 10 m/2.after 2sec.it reached the ground what is the height of the building

Answer 1

Answer

Height of the building = 20 m

Given

A stone is dropped from the top of a building.it's acceleration is 10 m/s². After 2 sec , it reached the ground

To Find

Height of the building

Concept Used

As the acceleration of the stone is constant throughout the motion . [ ∵ Gravity is always a constant ] So , we need to apply equation's of motion ,

→ v = u + at

→ s = ut + ¹/₂ at²

→ v² - u² = 2as

Solution

Initial velocity , u = 0 m/s

Since , dropped from top a building

Acceleration due to gravity , a = 10 m/s²

Time , t = 2 s

Displacement , s = ? m

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Apply 1st equation of motion ,

⇒ v = u + at

⇒ v = (0) + (10)(2)

⇒ v = 0 + 20

⇒ v = 20 m/s

So , Final velocity = 20 m/s

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Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (20)² - (0)² = 2(10)s

⇒ 400 - 0 = 20s

⇒ 20s = 400

⇒ $\rm 2s=40$

⇒ s = 20 m

So , Displacement of the stone = 20 m

Here , Displacement of the stone is equals to height of the building .

So , Height of the building = 20 m

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Answer 2

Answer :

Acc. due to gravity = 10m/s²

Time taken by stone to reach at the ground = 2s

We have to find height of the building$.$

ATQ, Acceleration due to gravity is said to be constant, we can apply equation of kinematics to solve this question.

For a body, Falling free under the action of gravity, g is taken positive and a body, Throwing upward against the action of gravity, g is taken negative$.$

Let's apply second eq. of kinematics,

➝ H = ut + (1/2) gt²

➝ H = (0×2) + (1/2) (10×2²)

➝ H = 0 + (1/2) (40)

➝ H = 20m

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