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Standard XII

Physics

1st Equation of Motion

A ball after ...

Question

A ball after having fallen from rest under the influence of gravity for $6 s$ , crashes through a horizontal glass plate, thereby losing two-third of its velocity. Then it reaches the ground in $2 s$ , height of the plate above the ground is

19.6 m

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39.2 m

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58.8 m

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78.4 m

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Solution

The correct option is C $58.8 m$
$v_{B} = v_{A} + g t = 0 + g (6) = 6 g$
Now, ${\frac{2}{3}}^{r d}$ of the velocity is lost.
Hence, $\frac{1}{3} \times 6 g = 2 g$ velocity is left.
$h = u t + \frac{1}{2} g t = (2 g) (2) + \frac{1}{2} g (2)^{2} = 6 g$ $= 58.8 m$

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A ball after having fallen from rest under the influence of gravity for 6s, crashes through a horizontal glass plate, thereby losing two-third of its velocity. Then it reaches the ground in 2s, height of the plate above the ground is

The correct option is C 58.8mvB=vA+gt=0+g(6)=6gNow, 23rd of the velocity is lost.Hence, 13×6g=2g velocity is left.h=ut+12gt=(2g)(2)+12g(2)2=6g=58.8 m

A ball after having fallen from rest under the influence of gravity for $6 s$ , crashes through a horizontal glass plate, thereby losing two-third of its velocity. Then it reaches the ground in $2 s$ , height of the plate above the ground is

The correct option is C $58.8 m$
$v_{B} = v_{A} + g t = 0 + g (6) = 6 g$
Now, ${\frac{2}{3}}^{r d}$ of the velocity is lost.
Hence, $\frac{1}{3} \times 6 g = 2 g$ velocity is left.
$h = u t + \frac{1}{2} g t = (2 g) (2) + \frac{1}{2} g (2)^{2} = 6 g$ $= 58.8 m$