E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If
it takes 0.6 seconds for it to reach the floor,
a. What is the vertical component of the ball's velocity just
before it hits the floor? (Use g = 10 m/s2.)
b. What is the horizontal component of the ball's velocity
just before it hits the floor?

Respuesta :

a) Vertical velocity: 5.9 m/s

b) Horizontal velocity: 5 m/s

Explanation:

a)

The motion of the ball is the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction)

- A uniformly accelerated motion (constant acceleration) along the vertical direction

Here we want to find the vertical component of the ball's velocity. This can be done by using the suvat equation for the vertical motion:

[tex]v_y = u_y +gt[/tex]

where:

[tex]v_y[/tex] is the vertical velocity at time t

[tex]u_y=0[/tex] is the initial vertical velocity (zero because the ball has been thrown horizontally)

[tex]g=10 m/s^2[/tex] is the acceleration of gravity (here we take downward as positive direction)

Substituting t = 0.6 s, which is the total time of flight, we find the vertical velocity of the ball just before it hits the ground:

[tex]v_y=0+(9.8)(0.6)=5.9 m/s[/tex]

b)

The motion along the vertical direction is an accelerated motion, because there is a force (the force of gravity) acting on the ball and that it causes an acceleration in the ball.

However, there are no forces acting in the horizontal direction on the ball (if we neglect the air resistance): this means that the acceleration of the ball in the horizontal direction is zero.

As a consequence, this also means that the horizontal component of the ball's velocity is constant during the motion.

Since the ball was thrown from the table with an initial horizontal velocity of 5 m/s, this means that the horizontal velocity of the ball just before it hits the floor is still

[tex]v_x = 5 m/s[/tex]

  • The vertical component of the ball's velocity just  before it hits the floor is 6.0m/s
  • The horizontal component of the ball's velocity  just before it hits the floor is 5m/s

a) To get the vertical component of the ball's velocity just  before it hits the floor, we will use the equation of motion expressed as:

[tex]v_y = u + gt[/tex]

where:

u is the initial velocity

g is the acceleration due to gravity

t is the time taken

Given the following:

  • u = 0m/s
  • g = 10m/s²
  • t = 0.6secs

Substitute into the formula as shown:

[tex]v_y=0+10(0.6)\\v_y=6.0m/s[/tex]

Hence the vertical component of the ball's velocity just  before it hits the floor is 6.0m/s

b) Since the ball rolls off a table with a horizontal velocity of 5 m/s, hence  the horizontal component of the ball's velocity  just before it hits the floor

is 5m/s

Learn more on velocity here: https://brainly.com/question/1125420

ACCESS MORE