Question

A body falls from the top of a building and reaches the ground 2.5seconds later. How high is the building??

Answer 1

Answer:

31.25 metres

Explanation:

Given :

• Time taken = t = 2.5 seconds

• Acceleration due to gravity = g = 10 m/s²

• Initial velocity = u = 0 m/s

To find :

• Height of the building

Using the second equation of motion :

S=ut+½×gt²

S=0×2.5+½×10×2.5²

S=0+5×6.25

S=31.25 metres

The height of the building is equal to 31.25 metres

More :

• First equation of motion = v=u+at

• Third equation of motion = v²-u²=2as

Answer 2

Given:

• Body falls from top of building
• Reaches ground 2.5 seconds later

To find:

• Height of building

Solution:

Here, the body falls downwards so , acceleration that means acceleration due to gravity = 9.8 m/s² and here initial velocity = 0 m/s because the body starts from rest

Now , finding height using the second equation of motion

S = ut + 1/2 at²

Where

• S : height of building = ?
• u : initial velocity = 0m/s
• t : time taken = 2.5s
• a : acceleration = 9.8 m/s² ≈ 10m/s²

Substituting the values we have

→ S = 0(2.5) + 1/2 10(2.5)²

→ S = 0 + 1/2 × 10 × 6.25

→ S = 1/2 × 62.5

→ S = 31.25 m

Hence,. height of building = 31.25 m