Question

If an object is dropped from the top of a building and it reaches the ground at t=4s, then the height of the building is ( ignoring air resistance) (g= 9.8 ms-2)

Answer 1

T= √(2h/g)
4^2=2h/g

h=78.4

I hope it helps

Answer 2

Answer:

The height of the building is 78.4 m.

Explanation:

Given that,

Time t = 4 s

When an object is dropped from the top of a building and it reaches the ground.

Then the initial velocity u = 0

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$...(I)

Where, u = initial velocity

g = acceleration due to gravity

t = time

Put the value of t and g in equation (I)

$s=0+\dfrac{1}{2}\times9.8\times4\times4$

$s = 78.4\ m$

Hence, The height of the building is 78.4 m.