Question

a stone is dropped from the roof of a building takes 4sec . to reach the ground the height of building is

Answer 1

Answer
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From Second Equation of Motion,
We can derive

$t = \frac{\sqrt{2h}}{\sqrt{g}}$

We have t = 4 sec
Taking g = 10 m/s²

=> h = t²g / 2

= h = 4² × 10 ÷ 2

=> h = 160 ÷ 2

=> h = 80 m
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HOPE IT HELPS :):):):)

Answer 2

$\bf{\underline{\underline{Answer:}}}$

$\bold{\pink{\green {\sf{\therefore\:Height\:=\:78.4\:m}}}}$

$\bf{\underline {Given:}}$

$⟹Initial \:velocity\: of \:object, \:u\: =\: 0$

$⟹Acceleration \:due\: to \:gravity, \:g = 9.8\: m\: {s}^{-2}$

$⟹Time, \:t = 4\: s$

$\bf{\underline {To\:Find:}}$

$⟹ The\: height\: of \:building = ?$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

We know that,

${h} = ut + \dfrac{1}{2}gt^2$

Substituting the values, we get

$h = \dfrac{1}{2}×9.8×16 = 78.4m$

$\bf{ Thus, \:the \:height \:of \:building\: is\: 78.4 \:m}$