Question

A body falls from the top of the building and reaches the ground 2s later.how high is the building

Answer 1

We will use newton's 2nd equation of motion i.e;

S=ut+1/2(a* t²)

here, u is initial velocity of object,

t is time taken,

a is acceleration of the object ,

And lastly S is the distance covered.

As given in the question now, the body falls freely so u is 0. Now, acceleration due to gravity will be the acceleration of the object and that is 9.8m/sec².Time taken is given and that is t=2.5s.

Applying the equation we get,

S= 1/2(a*t²)

(Since u is 0 )

so putting the values we get,

S= 1/2(9.8*2.5²)

S= 4.9(6.25)

S= 30.625 m

So we get our answer. The height of the building is 30.625 meters.