a block of mass slides off the edge of a 2 foot tall table at 3.13m/sec. how far away from the edge of the table does it land?
To determine how far away from the edge of the table the block lands, we can apply the principles of projectile motion.
Let's break down the problem step by step:
Step 1: Identify the given information:
- The height of the table: 2 feet
- The initial speed of the block: 3.13 m/s
Step 2: Determine the time it takes for the block to fall:
We can use the kinematic equation for vertical motion:
h = (1/2) * g * t^2, where h is the height of the table, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time of fall.
By substituting the given values, we can solve for t:
2 feet = (1/2) * 9.8 m/s^2 * t^2
Converting the height to meters:
0.61 meters = (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation:
t^2 = (0.61 meters * 2) / 9.8 m/s^2
t^2 = 0.1224 seconds^2
t ≈ √(0.1224) seconds
Step 3: Calculate the horizontal distance traveled by the block:
The horizontal distance can be determined using the equation:
d = v * t, where d is the horizontal distance, v is the horizontal velocity (which is constant), and t is the time of fall.
Given:
v = 3.13 m/s (initial speed)
t ≈ √(0.1224) seconds (time of fall)
Substituting the values, we can solve for d:
d = (3.13 m/s) * √(0.1224 seconds^2)
d ≈ 1.17 meters
Therefore, the block lands approximately 1.17 meters away from the edge of the table.