A student throws a ball vertically upward such that it travels 7.4m to its maximum height.
Part A
If the ball is caught at the initial height 2.7s after being thrown, what is the ball’s average speed?
Part B
What is the ball’s average velocity?
(B) since the ball ended up where it started, its average velocity was 0
Steve you may be correct, but both questions can be worded a little bit better. For A: I wasn't sure if d=7.4m or d=2(7.4)=14.8m. If so, then r=14.8m/2.7s=5.5m/s. For B: I wasn't sure if they wanted to know the average velocity for the trip up, or the trip up and back down. If so, then the average velocity is 0.
Part A
d=r*t
distance=rate*time
7.4m=r*2.7s
7.4m/2.7s=r
r=2.7m/s
Part B
Use the following equation:
Vf^2=Vi^2 +2 ad
Where:
Vf=0
a=g=9.8m/s
Vi=?
d=7.4m
0=Vi^2 +2(-9.8m/s^2)*(7.4m)
0=Vi^2-145m^2/s^2
145m^2/s^2=Vi^2
sqrt*(145)=Vi
Vi=12.0 m/s
AverageV=(Vf+Vi)/2=(0+12.0m/s)/2=6m/s
Yeah. I've noticed that either textbooks have gotten very sloppy in their language, or students who post here aren't very careful in transcribing their assignments.
Yes, but the for A the speed is 5.5m/s and 0 for the velocity. I must have been tired. I just didn' t read the question carefully.
Part A:
To find the average speed of the ball, we need to divide the total distance traveled by the time it took. In this case, the ball traveled a distance of 7.4m to its maximum height, and it took 2.7 seconds to reach the initial height.
Average speed = Total distance / Time taken
Average speed = 7.4m / 2.7s
Therefore, the ball's average speed is approximately 2.74 m/s.
Part B:
To calculate the average velocity, we need to consider both the magnitude and direction of motion. Since the ball was thrown vertically upwards and caught at the same height, the displacement (change in position) is zero. Hence, the average velocity is also zero.
In this case, the direction of motion changed, but the net displacement is zero since the ball ended at the same height from where it was thrown. Therefore, the ball's average velocity is 0 m/s.