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

Options

A. cannot hit the monkey
B. may hit the monkey if its weight is more than `30kg`
C. may hit the monkey if its weight is less than `30kg`
D. hits the monkey irrespective of its weight

Answer

D

Explanation

Let the monkey stationed at (A) be fired with a hun from (O) with a velocity (u) at an angle `theta` with the horizontal `OX` Draw `AC` perpendicular to `OX`~ Let the bullet cross the vertical line `AC` at (B) after time (t) and coordinates of (B) be (x, y) w.r.t origin (O)
.
:. `(OC)/(u cos theta) =x/(u cos theta)` ...(i)
(where OC=x)
In `Delta OAC`, `AC tan theta =x tan theta `...(ii)
Clearly `CB=y`=the vertical distance travelled by bullet in time (t).
Taking motion of the bullet from `O` to `B` along `Y`-axis we have
`y_(0) =0, y=y, u_(y) =u sin theta, a_(y) =- g, t=t`
As, `y=y_(0) + u_(y) t + 1/2 a_(y) t^(2)`
:. `y= 0` using theta `t + 1/2 (-g) t^(2)`
`= us sin theta t -1.2 gt^(2)`
:. `AB=AC-BC =x tan theta -y`
`=tan the -(u sin theta t 1/2 gt^(2))`
`=x tan theta -(u sin theta xx x/( u cos theta -1/2 gt2)`
[from (i)]
`=x tan theta -x tan theta + 1/2 gt ^(2) =1/2 gt^(2)`
It means the bullet will pass throughout the point
(B) on vertical line (AV) at a vertical distance `1/2 gt^(2)`
below point (A).
The distance through which the monkey falls
vertically in time `t =1/2 gt^(2) =AB`
It means the bullet and monkey will pass through the point (B) simultaneously. Therefore, the bullet will hit the monkey.