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Standard XII

Physics

1st Equation of Motion

A ball is dro...

Question

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. Time of travel above the window is (Take $g = 10 m / s^{2}$ )

0.5 s

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0.25 s

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0.35 s

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0.75 s

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Solution

The correct option is C 0.35 s
$s = u t + \frac{1}{2} g t^{2}$
$3 = u (0.5) + \frac{1}{2} \times 10 (0.5)^{2}$
$3 = (0.5) u + 5 (0.25)$
$\frac{1.75}{0.5} = u$
$\Rightarrow u = 3.5 m / s$
$v_{1} = u = 3.5 m / s$
$v_{1} = u_{1} + g t$
$3.5 = 10 (t) \Rightarrow t = 0.35 s e c$

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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. Time of travel above the window is (Take g=10m/s2)

The correct option is C 0.35 ss=ut+12gt23=u(0.5)+12×10(0.5)23=(0.5)u+5(0.25)1.750.5=u⇒u=3.5 m/sv1=u=3.5m/sv1=u1+gt3.5=10(t)⇒t=0.35 sec

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. Time of travel above the window is (Take $g = 10 m / s^{2}$ )

The correct option is C 0.35 s
$s = u t + \frac{1}{2} g t^{2}$
$3 = u (0.5) + \frac{1}{2} \times 10 (0.5)^{2}$
$3 = (0.5) u + 5 (0.25)$
$\frac{1.75}{0.5} = u$
$\Rightarrow u = 3.5 m / s$
$v_{1} = u = 3.5 m / s$
$v_{1} = u_{1} + g t$
$3.5 = 10 (t) \Rightarrow t = 0.35 s e c$