If a body dropped freely from a height of 20 | vedclass

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(C) Given: $h = 20 \,m$, $v = 31.4 \,ms^{-1}$.Using the equation of motion $v^2 = u^2 + 2gh$ where $u = 0$:$v^2 = 2gh$$g = \frac{v^2}{2h} = \frac{31.4

Vedclass · Accepted Answer

$2.5$

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(C) Given: $h = 20 \,m$, $v = 31.4 \,ms^{-1}$.Using the equation of motion $v^2 = u^2 + 2gh$ where $u = 0$:$v^2 = 2gh$$g = \frac{v^2}{2h} = \frac{31.4 	imes 31.4}{2 	imes 20} = \frac{985.96}{40} \approx 24.649 \,ms^{-2}$.Note that $31.4 \approx 10\pi$, so $g = \frac{(10\pi)^2}{40} = \frac{100\pi^2}{40} = 2.5\pi^2 \,ms^{-2}$.$A$ seconds pendulum has a time period $T = 2 \,s$.The formula for the time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$.Squaring both sides: $T^2 = 4\pi^2 \frac{l}{g} \Rightarrow l = \frac{T^2 g}{4\pi^2}$.Substituting $T = 2 \,s$ and $g = 2.5\pi^2 \,ms^{-2}$:$l = \frac{2^2 	imes 2.5\pi^2}{4\pi^2} = \frac{4 	imes 2.5\pi^2}{4\pi^2} = 2.5 \,m$.

If a body dropped freely from a height of $20 \,m$ reaches the surface of a planet with a velocity of $31.4 \,ms^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is (in $\,m$)

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