Question
A table tennis ball is hit vertically upwards with speed 20m/s. a) At what time does the ball have displacement of 3m? b) Hence calculate for how long the ball will be above a height of 3m. Show all working.
Answers
GPT 3.5
To find the time it takes for the ball to have a displacement of 3m, we can use the equation for vertical motion with constant acceleration:
s = ut + (1/2)at^2
Where:
s = displacement (3m)
u = initial velocity (20m/s)
a = acceleration (acceleration due to gravity = -9.8m/s^2)
t = time
Rearranging the equation to solve for time:
t = (2s/u)^0.5
Substituting the given values:
t = (2*3/20)^0.5
t = (6/20)^0.5
t = (0.3)^0.5
t ≈ 0.547s
Thus, the ball has a displacement of 3m at approximately 0.547s.
To calculate how long the ball will be above a height of 3m, we need to consider the path of the ball when it's going up and coming down.
When the ball is going up, the displacement from the starting point (height = 0) to the maximum height before it starts coming down is 3m. So the time it takes for the ball to go up and reach a height of 3m is equal to the time calculated above, which is approximately 0.547s.
When the ball is coming down, it will take the same amount of time to reach a height of 3m from the maximum height before it starts coming down. This is because the time it takes for the ball to reach its maximum height is symmetric to the time it takes for the ball to fall from its maximum height to the starting point.
Therefore, the ball will be above a height of 3m for approximately 2 * 0.547s = 1.094s.
s = ut + (1/2)at^2
Where:
s = displacement (3m)
u = initial velocity (20m/s)
a = acceleration (acceleration due to gravity = -9.8m/s^2)
t = time
Rearranging the equation to solve for time:
t = (2s/u)^0.5
Substituting the given values:
t = (2*3/20)^0.5
t = (6/20)^0.5
t = (0.3)^0.5
t ≈ 0.547s
Thus, the ball has a displacement of 3m at approximately 0.547s.
To calculate how long the ball will be above a height of 3m, we need to consider the path of the ball when it's going up and coming down.
When the ball is going up, the displacement from the starting point (height = 0) to the maximum height before it starts coming down is 3m. So the time it takes for the ball to go up and reach a height of 3m is equal to the time calculated above, which is approximately 0.547s.
When the ball is coming down, it will take the same amount of time to reach a height of 3m from the maximum height before it starts coming down. This is because the time it takes for the ball to reach its maximum height is symmetric to the time it takes for the ball to fall from its maximum height to the starting point.
Therefore, the ball will be above a height of 3m for approximately 2 * 0.547s = 1.094s.