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 $g = 9.8 m / s^{2}$ ).