# A stone is dropped vertically from the top of a tower of height 40 m. At the same time a gun is aimed directly at the stone from the ground at a horizontal distance 30 m from the base of the...

A stone is dropped vertically from the top of a tower of height 40 m. At the same time a gun is aimed directly at the stone from the ground at a horizontal distance 30 m from the base of the tower and fired. If the bullet from the gun is to hit the stone before it reaches the ground, what is the minimum velocity of the bullet must be, approximately.

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The stone drops down due to gravity, while the bullet travels due to its firing burst. The greater the velocity of the bullet the earlier (and higher up) it hits the stone. Lowest and latest, it can hit the stone at the moment the latter touches the ground.

Time taken by the stone to hit the ground can be obtained from the laws of motion.

`40 = 0*t+1/2 *9.81*t^2`

`rArr t = 2.855686 s`

If the stone is to be hit at the latest instance (which is the lowest point too), the bullet has to travel at least 30 m by this time (assuming the trajectory of the bullet to be linear in short distance).

Therefore the required minimum velocity of the bullet = 30/2.855686 = **10.51 m/s**.