A student throws a ball vertically upward such that it travels 9.0 m to its maximum height. If the ball is caught at the initial height 2.4s after being thrown, what is the ball's average speed?
I know average speed is change in distance over change in time, but I don't know how to get those numbers.
Thank you.
average speed= distance/time= 18m/time
Thank you very much.
To calculate the average speed, we need to find the change in distance and the change in time. Let's break down the problem step by step:
1. We are given that the ball traveled 9.0 m to its maximum height. This means that the distance the ball covered from the initial position to the maximum height is 9.0 m.
2. The ball is caught at the initial height 2.4s after being thrown. This means that the time it took for the ball to return to the initial height is 2.4 seconds.
To find the average speed, we can use the formula:
Average Speed = Change in Distance / Change in Time
In this case, the change in distance is the total distance covered by the ball, which is twice the distance to the maximum height (since it goes up and then comes down). So, the change in distance is 2 * 9.0 m = 18.0 m.
The change in time is the time it takes for the ball to go up to the maximum height and then come back down. Since the ball is caught 2.4 seconds after being thrown, the total time for the ball to reach the maximum height and return to the initial height is twice that, which is 2 * 2.4 s = 4.8 s.
Now, we can substitute the values into the formula:
Average Speed = 18.0 m / 4.8 s = 3.75 m/s
Therefore, the average speed of the ball is 3.75 m/s.