**Solution:**

To solve this problem, we can analyze the motion of the ball before and after the collision separately.

**Before Collision:**
- The ball falls from a height, which means it has an initial velocity of 0 m/s when it leaves the hand.
- Due to the effect of gravity, the ball accelerates downward at a rate of 9.8 m/s^2.
- The time taken by the ball to reach the floor can be calculated using the formula:
*s = ut + (1/2)at^2*, where
- *s* = distance traveled (height of the lift)
- *u* = initial velocity (0 m/s)
- *a* = acceleration (-9.8 m/s^2)
- *t* = time taken
- Substituting the given values, we can calculate the time taken:
*10 = 0*t + (1/2)(-9.8)t^2*
Solving this equation, we get *t = 1.43 s*.
- Therefore, the time taken by the ball to reach the floor is 1.43 seconds.

**During Collision:**
- The ball collides with the floor of the lift, which is moving upward with a velocity of 1 m/s.
- Since the collision is elastic, the total mechanical energy of the ball is conserved.
- The velocity of the ball just before the collision is 10 m/s downward.
- The velocity of the floor of the lift is 1 m/s upward.
- After the collision, the ball rebounds with a velocity in the opposite direction.
- The magnitude of the velocity after the collision can be calculated using the principle of conservation of mechanical energy:
*Initial kinetic energy + Initial potential energy = Final kinetic energy + Final potential energy*.
- The initial kinetic energy is given by (1/2)mv^2, where
- *m* = mass of the ball (assumed to be constant)
- *v* = velocity of the ball just before the collision
- The initial potential energy is given by mgh, where
- *g* = acceleration due to gravity (9.8 m/s^2)
- *h* = height of the lift
- The final kinetic energy is given by (1/2)mv'^2, where
- *v'* = velocity of the ball after the collision
- The final potential energy is given by mgh', where
- *h'* = height of the lift after the collision (same as initial height)
- Since the mass and height remain constant, the equation can be simplified to:
*(1/2)v^2 + mgh = (1/2)v'^2 + mgh'*
- Substituting the given values:
*(1/2)(10^2) + m(9.8)(10) = (1/2)(v'^2) + m(9.8)(10)*
*500 = (1/2)(v'^2) + 980m*
- Rearranging the equation, we get:
*(1/2)(v'^2) = 500 - 980m*
*v'^2 = 1000 - 1960m*
- Since the mass of the ball does not change, we