a child knocks a ball off a table with a height of 0.60 m. The initial velocity of the ball is 2.4 m/s. What will be the horizontal distance between the edge of the table and where the ball lands?
To find the horizontal distance between the edge of the table and where the ball lands, we can use the equations of motion for projectile motion. These equations describe the motion of an object that is projected into the air with an initial velocity.
First, let's analyze the vertical motion of the ball. The ball is initially at a height of 0.60 m above the ground, and it will reach the ground (final height = 0 m). The acceleration due to gravity can be considered constant and equal to 9.8 m/s².
Using the equation:
h = ut + (1/2)gt²
where:
- h is the vertical distance traveled by the ball (0 m),
- u is the initial vertical velocity (0 m/s since the ball is projected horizontally),
- t is the time of flight,
- g is the acceleration due to gravity (9.8 m/s²),
We can rearrange the equation to solve for t:
0 = 0 + (1/2)(9.8)t²
0 = 4.9t²
t² = 0
t = 0
Since the time of flight (t) is 0, it means that the ball will fall vertically from the edge of the table without any horizontal displacement. Therefore, the horizontal distance between the edge of the table and where the ball lands will be zero.
Note: It's assumed that there are no external factors affecting the motion of the ball, such as air resistance.