100 kg gun fires a ball of 1 kg rice in delhi from a cliff of height 5

1.	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» Given : m1 = 100 kg m2 = 1 kg h = 500 m x = 400 m To Find : RECOIL velocity of the gun Solution : Let u be velocity of ball and t be TIME by the ball to hit the ground. $\large \tt h = \dfrac{1}{2} {gt}^{2} \\ \\ \implies\tt 500 = \frac{1}{2} 10 {t}^{2} \\ \\ \tt\implies t = \sqrt{100} \\ \\ \tt\implies t = 10 \: sec$ ____________________________ $\large\tt x = ut \\ \\\implies \tt u = \frac{x}{t} \\ \\ \implies \tt u = \frac{400}{10} \\ \\ \implies \tt u = 40 \: m \: {s}^{ - 1}$ If v is recoil velocity of gun , then according to principle of CONSERVATION of linear MOMENTUM $\large\tt m_1v =m_2u \\ \\\implies \tt v = \frac{m_2u}{m_1} \\ \\\implies \tt v = \frac{1}{100} \times 40 \\ \\ \large \boxed{\tt \green{ v = 0.4 \: m \: {s}^{ - 1}}}$ Recoil velocity of gun is 0.4 m/s

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»

Given :

m1 = 100 kg
m2 = 1 kg
h = 500 m
x = 400 m

To Find :

RECOIL velocity of the gun

Solution :

Let u be velocity of ball and t be TIME by the ball to hit the ground.

$\large \tt h = \dfrac{1}{2} {gt}^{2} \\ \\ \implies\tt 500 = \frac{1}{2} 10 {t}^{2} \\ \\ \tt\implies t = \sqrt{100} \\ \\ \tt\implies t = 10 \: sec$

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$\large\tt x = ut \\ \\\implies \tt u = \frac{x}{t} \\ \\ \implies \tt u = \frac{400}{10} \\ \\ \implies \tt u = 40 \: m \: {s}^{ - 1}$

If v is recoil velocity of gun , then according to principle of CONSERVATION of linear MOMENTUM

$\large\tt m_1v =m_2u \\ \\\implies \tt v = \frac{m_2u}{m_1} \\ \\\implies \tt v = \frac{1}{100} \times 40 \\ \\ \large \boxed{\tt \green{ v = 0.4 \: m \: {s}^{ - 1}}}$

Recoil velocity of gun is 0.4 m/s