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Standard XII

Physics

Projectile from a Height

A boy standin...

Question

A boy standing on the viewing platform of the Qutub Minar. He then throws a ball vertically downwards at $2 m/s$ . If the platform is $72 m$ above ground, the speed with which the ball hits the ground in $(m/s)$ is

48 m/s

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38 m/s

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40 m/s

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34 m/s

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Solution

The correct option is B $38 m/s$
As the initial velocity, $u = 2 m/s$ , is along the acceleration due to gravity, we have
$v^{2} = u^{2} + 2 a s$
$v = \sqrt{(2)^{2} + 2 (10) (72)}$
$= \sqrt{1444}$
$v = 38 m/s$

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A boy standing on the viewing platform of the Qutub Minar. He then throws a ball vertically downwards at 2 m/s. If the platform is 72 m above ground, the speed with which the ball hits the ground in (m/s) is

The correct option is B 38 m/sAs the initial velocity, u=2 m/s, is along the acceleration due to gravity, we have v2=u2+2as v=√(2)2+2(10)(72) =√1444 v=38 m/s

A boy standing on the viewing platform of the Qutub Minar. He then throws a ball vertically downwards at $2 m/s$ . If the platform is $72 m$ above ground, the speed with which the ball hits the ground in $(m/s)$ is

The correct option is B $38 m/s$
As the initial velocity, $u = 2 m/s$ , is along the acceleration due to gravity, we have
$v^{2} = u^{2} + 2 a s$
$v = \sqrt{(2)^{2} + 2 (10) (72)}$
$= \sqrt{1444}$
$v = 38 m/s$