Question

A person takes an aim at a monkey sitting on a tree and fires a bullet. Seeing the smoke begins to fall freely then the bullet will

• A. Always hit the monkey
• B. Go above the monkey
• C. Go below the monkey
• D. Hit the monkey if the initial velocity of the bullet is more than a certain velocity

Answer 1

Answer: (D)

Hit the monkey if the initial velocity of the bullet is more than a certain velocity

If there were no gravity, the bullet would reach height $H$ in time $t$ taken by it to travel the horizontal distance $X$, i.e., $H=u\sin \theta t$ and $x=u\cos \theta t$ or $t=\frac{x}{u\cos \theta }$
However, because of gravity the bullet has an acceleration $g$ vertically downwards, so in time $t$ the bullet will reach a height
$y=u\sin \theta \times t-\frac{1}{2}g{t}^2=H-\frac{1}{2}g{t}^2$
This is lower than $H$ by $\frac{1}{2}g{t}^2$, which is exactly the amount the monkey falls in this time. So the bullet will hit the monkey regardless of the initial velocity of the bullet so long as it is great enough to travel the horizontal distance to the tree before hitting the ground. (For large $u$ lesser will be the time of motion; so the monkey is hit near its initial position and for smaller $u$ it is hit just before it reaches the floor). Bullet will hit the monkey only and only if
$y>0$ i.e., $H-\frac{1}{2}g{t}^2>0$
or $H>\frac{1}{2}g{t}^2$ or $H>\frac{1}{2}g\frac{x^2}{u^2{\cos }^2\theta }$
or $u>\frac{x}{\cos \theta }\sqrt{\frac{g}{2H}}$ or $u>\sqrt{\frac{g}{2H}(x^2+H^2)}=u_0$
If $u<u_0$, the bullet will hit the ground before reaching the monkey.