Calculating the Time Difference for Two Projectiles in Motion
Understanding the Scenario
Two students are on a balcony 19.6m above the street. One student throws a ball vertically downward at 14.7m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just missed the balcony on the way down.
The Question
What is the difference in the time the ball spend in the air?
Breaking Down the Solution
For the ball going straight down, it takes 1 second. For the ball going up, it takes 3 seconds. Therefore, the final answer is: 3 seconds.
Explanation
The question involves calculating the time difference for two balls thrown vertically, one going downward and the other going upward from the same height and speed. The ball thrown downward takes 1 second to reach the street, while the ball thrown upward takes 3 seconds to return to the starting point.
Since the upward thrown ball has to come back to the street from the same height, it will take another 1 second. Thus, the total time for the upward thrown ball is 3 seconds (going up and returning) + 1 second (going down to the street) = 4 seconds.
Therefore, the time difference in the air between the two balls is 4 seconds - 1 second = 3 seconds.
Two students are on a balcony 19.6m above the street. One student throws a ball vertically downward at 14.7m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just missed the balcony on the way down. What is the difference in the time the ball spend in the air? The difference in time two balls spend in the air, one thrown vertically downward and the other upward but from the same height and speed is 3 seconds, with the upward thrown ball spending more time in the air.