Let's solve the problem step by step.

Given:

• Initial horizontal velocity ($v_{x0}$) = 3 m/s
• Time to reach the floor ($t$) = 0.4 seconds
• Acceleration due to gravity ($g$) = 10 m/s²

Part (a): Downward Component of the Ball’s Velocity Just Before It Hits the Floor

The downward component of the velocity is influenced by gravity. The initial vertical velocity of the ball is zero since it rolls off the table horizontally. We can calculate the final vertical velocity ($v_y$) just before it hits the floor using the formula:

$v_y = g \cdot t$

Substitute the values:

$v_y = 10 \, \text{m/s}^2 \times 0.4 \, \text{s} = 4 \, \text{m/s}$

Answer to (a):

The downward component of the ball’s velocity just before it hits the floor is 4 m/s.

Part (b): Horizontal Component of the Ball’s Velocity Just Before It Hits the Floor

The horizontal component of the velocity ($v_x$) remains constant throughout the motion since there is no horizontal acceleration. Therefore, it is equal to the initial horizontal velocity:

$v_x = v_{x0} = 3 \, \text{m/s}$

Answer to (b):

The horizontal component of the ball’s velocity just before it hits the floor is 3 m/s.

Summary:

• (a) Downward velocity component: 4 m/s
• (b) Horizontal velocity component: 3 m/s

Would you like more details or have any questions?

Related Questions:

1. What would be the total velocity of the ball just before it hits the floor?
2. How far horizontally does the ball travel before hitting the ground?
3. What is the kinetic energy of the ball just before it hits the floor?
4. How would the downward component of velocity change if the time to reach the floor doubled?
5. What would be the impact speed of the ball if it fell from a greater height?

Tip:

To find the total velocity of an object moving in two dimensions, combine the horizontal and vertical components using the Pythagorean theorem.