At a certain instant a ball passes location < 11, 10, -16 > m. In the next 3 seconds, the ball's average velocity is < -8, 37, 30 > m/s. At the end of this 3 second time interval, what is the height y of the ball?
-16 + 3 (30)
To find the height (y-coordinate) of the ball at the end of the 3-second time interval, we need to calculate the change in the y-coordinate and add it to the initial y-coordinate.
Given:
Initial position = < 11, 10, -16 > m
Average velocity = < -8, 37, 30 > m/s
Time interval = 3 seconds
Step 1: Calculate the change in position.
To find the change in position, we multiply the average velocity by the time interval:
Change in position = Average velocity * Time interval
= < -8, 37, 30 > m/s * 3 s
= < -24, 111, 90 > m
Step 2: Add the change in position to the initial position.
To find the final position, we add the change in position to the initial position:
Final position = Initial position + Change in position
= < 11, 10, -16 > m + < -24, 111, 90 > m
= < -13, 121, 74 > m
Step 3: Obtain the height (y-coordinate) of the ball.
The height (y-coordinate) of the final position is given by the y-component of the final position vector:
Final height (y-coordinate) = 121 m
Therefore, at the end of the 3-second time interval, the height (y-coordinate) of the ball is 121 meters.