A ball moving at +3.0 m/s along a table rolls off a table and lands on the ground 2.0 m away. How high was the table?
h = 9.8 × (2/3)² / 2 = 2.2 m
To find the height of the table, we can make use of the principles of projectile motion.
Given the initial velocity of the ball along the table is +3.0 m/s, and it lands 2.0 m away on the ground, we need to determine the height of the table.
We can break down the problem into two separate motions: the horizontal motion and the vertical motion.
First, let's consider the horizontal motion. The horizontal motion is uniform, meaning there is no acceleration in the horizontal direction. The time taken for the ball to travel 2.0 m horizontally can be calculated using the formula:
time = distance / velocity
time = 2.0 m / 3.0 m/s = 2/3 s
Now, let's consider the vertical motion of the ball. We need to determine the height of the table, which is the vertical distance traveled by the ball.
Using the kinematic equation:
h = (1/2) * g * t^2
where
h is the height,
g is the acceleration due to gravity (taken as 9.8 m/s^2),
t is the time taken for vertical motion (2/3 s).
Plugging in the values:
h = (1/2) * 9.8 m/s^2 * (2/3 s)^2
= (1/2) * 9.8 m/s^2 * 4/9 s^2
= (4.9/9) m^2/s^2
Therefore, the height of the table is approximately 2.2 meters.