A stone is dropped from a roof. How much time does it take to fall 4.9 m? (Takeg = 9.8 m/s2).

A stone is dropped from a roof of height

4.9 m

. The time taken to fall is

1 s .

As the stone is dropped, the initial velocity is

0 m / s

. The acceleration of the stone is the acceleration due to gravity

(g)

in the downward direction.
So, initial velocity

(u) = 0 m / s

Acceleration

(a) = 9.8 m / s^{2}

acceleration a will be equal to

g

(acceleration due to gravity) in the state of free-falling
Distance travelled

(s) = 4.9 m

Applying the equation of motion

s = ut + \frac{1}{2} a t^{2}

Putting the values

\Rightarrow 4.9 = 0 + \frac{1}{2} \times 9.8 \times t^{2}

{\Rightarrow t}_{} = \sqrt{\frac{2 \times 4.9}{9.8}}

{\Rightarrow t}_{} = 1 s

A stone is dropped from a roof. How much time does it take to fall $4.9 m$ ? (Take $g = 9.8 m / s^{2}$ ).