**Force exerted on the floor by the ball**

To determine the force exerted on the floor by the ball, we need to consider the concept of impulse. Impulse is defined as the product of force and the time over which it acts. It can be calculated using the formula:

Impulse (J) = Force (F) × Time (Δt)

In this case, the ball is dropped vertically and hits the floor with a certain speed. Upon hitting the floor, it rebounds with a different speed. The time for which the ball is in contact with the floor is given as 0.03 seconds.

**Calculating the impulse exerted on the ball**

To calculate the impulse exerted on the ball, we can use the concept of conservation of momentum. The momentum before the collision is equal to the momentum after the collision, assuming no external forces act on the system.

Momentum (p) = Mass (m) × Velocity (v)

Before the collision:
Initial momentum = (1.5 kg) × (25 m/s) = 37.5 kg·m/s

After the collision:
Final momentum = (1.5 kg) × (15 m/s) = 22.5 kg·m/s

The change in momentum (Δp) is given by:
Change in momentum = Final momentum - Initial momentum
Δp = 22.5 kg·m/s - 37.5 kg·m/s = -15 kg·m/s (negative sign indicates a change in direction)

Since impulse is equal to the change in momentum, we can calculate it as:
Impulse = -15 kg·m/s

**Calculating the force exerted on the ball**

Now, using the formula for impulse, we can calculate the force exerted on the ball by dividing the impulse by the time of contact:

Force = Impulse / Time

Force = (-15 kg·m/s) / (0.03 s)
Force = -500 N

Therefore, the force exerted on the floor by the ball is 500 N. The negative sign indicates that the force is acting in the opposite direction of the ball's motion.

2000NF=∆p÷∆t∆p=m(v-u) =1.5[25-(-15)] =60 kgm/sF=60÷0.03 =2000N