Question

$\huge\underline{\bf \orange{Question :}}$

100 kg gun fires a ball of 1 kg rice in delhi from a cliff of height 500 metre it falls on the ground at a distance of 400 metre from the bottom of the cliff the recoil velocity of the gun is ? ​

Answer 1

heya,

Here,

Mass of the gun, M = 100 kg

Mass of the ball, m= 1 kg

Height of the chiff, h= 500 m

g = 10ms

−2

Time taken by the ball to reach the ground is

t= √2g/h

= √2×500m/10m/s

Horizontal distance covered = ut

∴400=u×10

Where u is the velocity of the ball.

u=40 ms −1

According to law of conservation of linear momentum, we get,

0=Mv+mu

v=− -mu/M

1×40/100

=−0.4ms −1

-ve sign shows that the direction of recoil of the gun is opposite to that of the ball.

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Answer 2

$\huge\bold{\purple{\fbox {\pink{\mathbb{ANSWER}}}}}$

$\bold{\longmapsto}$Here,

Mass of the gun, M = 100 kg

Mass of the ball, m= 1 kg

Height of the chiff, h= 500 m

g = 10ms

−2

Time taken by the ball to reach the ground is

t= g 2h = 10ms −12×500m

Horizontal distance covered = ut

∴400=u×10

Where u is the velocity of the ball.

u=40 ms −1

According to law of conservation of linear momentum, we get,

0=Mv+mu v=− M mu =− 100kg (1kg)(40ms −1 ) =−0.4ms −1

-ve sign shows that the direction of recoil of the gun is opposite to that of the ball.

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$\bold{\purple{\fbox {\green{Hope\:it\:helps\:you}}}}$

$\bold{\purple{\fbox {\orange{Plz\:mark\:as\:brainliest}}}}$