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A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops. The bullet

Question :- A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops. The bullet –

(A) Cannot hit the monkey

(B) It will hit the monkey

(D) None of these

Solution :-

Time taken by bullet to reach point B,

$\displaystyle {{t}_{B}}=\frac{x}{u\cos \theta }$ …..(1)

Time taken by monkey to reach point B,

Using S = ut + (1/2) gt² we get,

$\displaystyle AB=0\times t+\frac{1}{2}gt_{M}^{2}$

$\displaystyle \Rightarrow {{t}_{M}}=\sqrt{\frac{2(AB)}{g}}$ …..(2)

Now using the equation of trajectory, i.e.,

$\displaystyle y=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$

We get,

$\displaystyle \Rightarrow BC=x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$

AB = AC – BC

$\displaystyle \Rightarrow AB=AC-\left[ x\tan \theta -\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta } \right]$

Here AC = x tanθ, so

$\displaystyle AB=x\tan \theta -x\tan \theta +\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$

$\displaystyle \Rightarrow AB=\frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta }$ …..(3)

Using the value of equation (3) in equation (2), we get

$\displaystyle {{t}_{M}}=\sqrt{\frac{2}{g}\left( \frac{g{{x}^{2}}}{2{{u}^{2}}{{\cos }^{2}}\theta } \right)}$

$\displaystyle \Rightarrow {{t}_{M}}=\frac{x}{u\cos \theta }={{t}_{B}}$ …..(4)

From equation (4) it is clear that the bullet and monkey will reach the point B in same time, hence the bullet will hit the monkey. So option (B) is correct.

A hunter aims his gun and fires a bullet directly at a monkey on a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops. The bullet

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