A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10 m / s 2 ) :

Question

A ball of mass 200 g rests on a vertical post of height 20 m. A bullet of mass 10 g, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m and the bullet at a distance of 120 m from the foot of the post. The value of initial velocity of the bullet will be (if g=10 m / s 2 ) : (A) 400 m / s (B) 360 m / s (C) 60 m / s (D) 120 m / s

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1bd43c19087e415c74755b5e9b36eaca-.png" /> v 1=(30/√(2 h ) g ), v 2=(120/√(2 h ) g ) (0.01) u =(0.2) (30 √ g /√2 h )+(0.01) (120 √ g /√2 h ) u =300+60=360 ms -1

$400 m / s$

$360 m / s$

$60 m / s$

$120 m / s$

$v_{1} = \frac{30}{\frac{2 h}{g}}, v_{2} = \frac{120}{\frac{2 h}{g}}$
$(0.01) u = (0.2) \frac{30 g}{2 h} + (0.01) \frac{120 g}{2 h}$
$u = 300 + 60 = 360 m s^{- 1}$

400m/s

360m/s

60m/s

120m/s

v1​=g2h​​30​,v2​=g2h​​120​ (0.01)u=(0.2)2h​30g​​+(0.01)2h​120g​​ u=300+60=360ms−1

$400 m / s$

$360 m / s$

$60 m / s$

$120 m / s$

$v_{1} = \frac{30}{\frac{2 h}{g}}, v_{2} = \frac{120}{\frac{2 h}{g}}$
$(0.01) u = (0.2) \frac{30 g}{2 h} + (0.01) \frac{120 g}{2 h}$
$u = 300 + 60 = 360 m s^{- 1}$