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  • A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun.

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun.

Answers (1)

Let u ms-1 be the horizontal speed of the ball & its vertical component will be zero. If we consider the motion of the ball vertically downward then,

u = 0, s = h = 500m, and g = 10sm-2,

We know that, s = ut + \frac{1}{2} at^{2}

500 = a \times t + \frac{1}{2} \times 10t^{2}

t = \sqrt{100}  = 10 s

Its horizontal range will be equal to u \times 100

400 = v \times 10

Thus, v = 40 m/s

Assume the mass and velocities of the ball and gun are given by,

m_{b} is the mass of the ball

M_{g} is the mass of the gun

u_{b} is the initial velocity of the ball

v_{g} is the final velocity of the gun

& v_{b} is the final velocity of the ball

Now, according to the figure, we get 

 m_{g}(0) + M_{G}(0) = 1(40) + 100v_{G}

Therefore, the recoil velocity of the gun will be -\frac{40}{100}ms^{-1} = -\frac{2}{5} ms^{-1} = -0.4 ms^{-1}, which is opposite to the speed of the ball.

Posted by

infoexpert24

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