a stone is thrown upward with a speed of 6m/s from the ground. a man who catches the stone is located 1 meters above the ground.?

kinematics

v² = u² + 2as

Taking acceleration due to gravity = 10 m/s² downwards

v² = 6² + 2 * -10 * 1

v = 4 m/s

Vo^2 = V^2+2gh

V = √6^2-19.6 = 4.05 m/sec

Does he catch it while the stone is going up, or while it's going down? Turns out for this problem, it doesn't make a difference: Going up, he catches it at a speed of 4.05 m/s (upwards), and going down, he catches it at a speed of -4.05 (downwards).

The problem is worded poorly.

We have to assume that the stone is somehow "thrown" from ground level.

- Maybe some guy is standing in a pit when he throws it

That means that the stone travels upward 1 m.

For these sorts of problems

the trick is to identify which variables you know

and

which variable you need

and then find the equation that fits that information

Variables known:

Vzero = 6 m/s

distance (X - Xzero) = 1 m

acceleration (A) = -9.8 m/s (negative because the force is in the opposite direction of V and X)

Variable needed:

Vfinal

These variables fit the motion equation

V-squared - Vzero-squared = 2A(X - Xzero)

just plug in the values and solve for V

(The correct answer is 4 m/s)