A bowling ball rolls off the edge of a giant' table at 18 m/s. If it takes 6 s for the ball to hit the ground.
A) How far does it land from the base of the table?
B) how far does it fall?
A) 18 m/s * 6 s
B) 1/2 g t^2
To determine how far the bowling ball lands from the base of the table, we can use the formula for horizontal distance traveled:
\( \text{{Horizontal distance}} = \text{{Initial horizontal velocity}} \times \text{{Time of flight}} \)
In this case, the initial horizontal velocity is 18 m/s and the time of flight is 6 seconds. Therefore:
\( \text{{Horizontal distance}} = 18 \, \text{m/s} \times 6 \, \text{s} = 108 \, \text{m} \)
So, the bowling ball lands 108 meters from the base of the table.
To determine how far the bowling ball falls, we can use the formula for vertical distance traveled under constant acceleration:
\( \text{{Vertical distance}} = \frac{1}{2} \times \text{{Acceleration}} \times \text{{Time of flight}}^2 \)
The acceleration due to gravity is approximately 9.8 m/s² (assuming no air resistance). Since the ball falls for 6 seconds, we can calculate:
\( \text{{Vertical distance}} = \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (6 \, \text{s})^2 = 176.4 \, \text{m} \)
Therefore, the bowling ball falls approximately 176.4 meters.