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. Will the bullet hit the monkey ?

Explanation

Horizontal distance travelled
`OB =x=ucos theta t` or `t=x/(u cos theta)`
For motion of bullet from O to B, the vertical height `AB=u sin thetat -1/2 "gt"^(2)`
`=u sin theta (x/(u cos theta)) -1/2"gt"^(2) =x tan theta -("gt"^(2))/2`…….(i)
Also from figure MB `=x tan theta`
Now, the height through which monkey falls
y=MA =MB -AB
`=x tan theta -(x tan theta -("gt"^(2))/2) =1/2 "gt"^(2)`

Thus, in time t the bullet passes through A a vertical distance `1/2"gt"^(2)` below M.
The vertical distance through which the monkey fall in time t. `s=1/2"gt"^(2)`
Thus, the bullet and the monkey will always reach at point A at the same time.