Question

pls solve ?...............​

Answer 1

Given :

• initial velocity = 0 m /s

( as the coin is dropped it's initial velocity will be zero )

• height of the coin from the floor of the elevator = 2.45 m

• acceleration due to gravity = 9.8m /s²

Solution :

By using the second kinematic equation ,

$\displaystyle\star \underline{ \boxed {\green{s = ut + \frac{1}{2} g {t}^{2} }}}$

here ,

• s = distance
• u = initial velocity
• t = time taken
• g = acceleration due to gravity

Substituting the given values in the above equation we get ,

$\displaystyle \implies{s = \frac{1}{2} g{t}^{2} }$

$\displaystyle \implies{{t}^{2} = \frac{2s}{ g} }$

$\displaystyle \implies{ {t}^{2} = \frac{2 \times 2.45}{9.8} }$

$\displaystyle \implies{ {t}^{2} = \frac{4.9}{9.8} }$

$\displaystyle \implies{{t}^{2} = \frac{1}{2} }$

$\displaystyle \boxed{ \red{t = \frac{1}{ \sqrt{2} } }}$

Answer 2

▪ $\underline\mathrm{Given-}$

• $\underline\mathrm{Initial \: Velocity (u) = 0}$
• $\underline\mathrm{Distance = 2.45}$
• $\underline\mathrm{Acceleration = 9.8}$

▪ $\underline\mathrm{To \: Find-}$

• $\underline\mathrm{Time = ?}$

▪ $\underline\mathrm{Solution-}$

We Know 2nd Kinematic / Eqn of Motion,

$★ \: s = ut + \frac{1}{2} \times {gt}^{2}$

Here,

• $\underline\mathrm{S = Distance}$
• $\underline\mathrm{U = Initial \: Velocity}$
• $\underline\mathrm{T = Time}$
• $\underline\mathrm{G = Acceleration (Gravity)}$

Placing Values in the Equation,

$⟶s = 0 \times 0 + \frac{1}{2} \times {gt}^{2}$

$⟶s = \frac{1}{2} \times {gt}^{2}$

$⟶ {t}^{2} = \frac{2s}{g}$

$⟶ {t}^{2} = \frac{2 \times 2.45}{9.8}$

$⟶ {t}^{2} = \frac{\cancel{4.9}}{\cancel{9.8}}$

$⟶ {t}^{2} = \frac{1}{2}$

$⟶t = \frac{1}{ \sqrt{2} }$

$\underline\mathrm{⛬ Option \: B \: Is \: Correct!}$