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A boy throws a stone vertically up to a man standing 2.0 m above the boy. lf thestone is thrown up with a velocity of 7.0 m/s, what is the velocity of the stone at theinstant when it is caught by the man? The acceleration due to free fall is '10 m/s2.

Question

A boy throws a stone vertically up to a man standing 2.0 m above the boy. lf thestone is thrown up with a velocity of 7.0 m/s, what is the velocity of the stone at theinstant when it is caught by the man? The acceleration due to free fall is '10 m/s2.

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Solution

To solve this problem, we can use the equation of motion:

v² = u² + 2as

where: v is the final velocity (which we want to find), u is the initial velocity (7.0 m/s in this case), a is the acceleration (which is -10 m/s² because it's against the direction of motion), s is the distance (2.0 m in this case).

Substituting the given values into the equation, we get:

v² = (7.0 m/s)² + 2*(-10 m/s²)*(2.0 m)

Solving this equation will give us the final velocity of the stone when it is caught by the man.

This problem has been solved

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