A20kg boy stands on a small stationary (at rest) merry-go-round near the edge of the merry-go-round. the total moment of inertia of the system of merry-go-round with the boy on it about the center is 120kg m2 . the boy at the edge of the merry-go-round (radius of 2m) jumps off the merry-go-round in a tangential direction with a liner speed of 1.5m/s. what is the angular speed of the merry-go-round after the boy leaves it?

The angular speed of the merry-go-round after the boy leaves it is 2.25 rad/s.

Explanation:

Given that,

Mass of boy = 20 kg

Total moment of inertia = 120 kgm²

Radius = 2 m

Speed = 1.5 m/s

We need to calculate the moment of inertia of boy

Put the value into the formula

We need to calculate the

Using conservation of angular momentum

Put the value into the formula

Hence, The angular speed of the merry-go-round after the boy leaves it is 2.25 rad/s.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

i think the answer is a and b

i wish you luck