Question

In a building a ball is dropped from u=0 from top of building one-fourth of distance covered by body in last 2 seconds find A) TOTAL TIME B) HEIGHT OF BUILDING

Answer 1

Answer

A ) 14.81 s

B ) 109.66 m

Given

• In a building a ball is dropped from u=0 from top of building one-fourth of distance covered by body in last 2 seconds

To Find

A) TOTAL TIME

B) HEIGHT OF BUILDING

Solution

Case - I : For whole distance , s = x

Here ,

Initial velocity , u = 0

Use 2nd equation of motion .

⇒ s = ¹/₂ gt²

⇒ x = ¹/₂ (10)t²

⇒ x = 5t² ... ( I )

Case - II : For 1st 3x/4 m

u = 0

s = 3x/4

t = t-2

⇒ s = ¹/₂ gt²

⇒ 3x/4 = 5(t-2)²

⇒ 3x = 20 (t-2)²

⇒ 3 ( 5t² ) = 20 (t-2)² [ From ( I ) ]

⇒ 3/4 t² = (t-2)²

⇒ √3/2 t = t-2

⇒ t (1-√3/2) = 2

⇒ t = 2/ [ (1-√3/2) ]

⇒ t ≅ 14.81 s

Now take ( I )

⇒ x = 5t²

⇒ x = 5(14.81)²

⇒ x = 109.66 m