A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the $3$ m length of a window at certain distance from the top of the building.
( $g = 10 m / s^{2}$ )
How far is the top of the window from the point at which the ball was dropped ?

0.5

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0.5225

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0.6125

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0.8

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Solution

The correct option is A $0.6125$ m

$s = u t + \frac{1}{2} g t^{2}$
$3 = u (0.5) + \frac{1}{2} \times 10 (0.5)^{2}$
$3 = (0.5) u + 5 (0.25)$
$\frac{1.75}{0.5} = u$
$\Rightarrow u = 3.5 m / s$
$u^{2} = u_{1}^{2} + 2 g s$
$\frac{(3.5)^{2}}{2 g} = s$
$\Rightarrow s = \frac{3.5 \times 3.5}{2 g}$ $\Rightarrow s = 0.6125 m$

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A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m length of a window at certain distance from the top of the building. (g=10m/s2)How far is the top of the window from the point at which the ball was dropped ?

The correct option is A 0.6125 m s=ut+12gt23=u(0.5)+12×10(0.5)23=(0.5)u+5(0.25)1.750.5=u⇒u=3.5 m/su2=u21+2gs(3.5)22g=s⇒s=3.5×3.52g⇒ s=0.6125 m

A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the $3$ m length of a window at certain distance from the top of the building.
( $g = 10 m / s^{2}$ )
How far is the top of the window from the point at which the ball was dropped ?