Question

A bomb is thrown at a speed 20 m/s at an angle 45°. At the highest point, it explodes into two parts of equal
mass, the one part coming to rest. The distance from the origin to the point where the other part strikes the
ground
(1) 60 m
(2) 40 m
(3) 20 m
(4) 10 m
​

Answer 1

Answer:

-580 because I feel so and it must be either this or 340

Answer 2

Answer:

A. 60 m

Explanation:

I hope you are clear on the subject of center of mass, as I will be using it for solving this question.

Across the process of blasting and the fragments of the bomb falling down, the center of mass of the bomb will still trace the same parabolic trajectory. That is, the range of the center of mass of the bomb will be the same.

Range of Center of mass (Rc) = 2u²sinθcosθ/g = 40 meters.

Now, at max height, Horizontal distance covered by the bomb is: 1/2 of Range = 20 meters. Therefore, the first part of the bomb is 20 meters away from the ground. Now using Center of Mass formula:

∑mx/∑m = Position of center of mass.

∴m/2 * 20 + m/2 * x     =       40

        m/2 + m/2

x = 60