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.
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.
To solve this problem, we can use the equations of motion for an object in free fall.
a) To find the time at which the ball has a displacement of 3m, we can use the equation:
s = ut + (1/2)at^2
Where:
s = displacement (3m)
u = initial velocity (20m/s)
a = acceleration due to gravity (-9.8m/s^2, negative sign as it acts downwards)
t = time
Let's substitute the given values into the equation:
3 = (20)t + (1/2)(-9.8)t^2
Rearranging the equation to quadratic form:
-4.9t^2 + 20t - 3 = 0
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = -4.9, b = 20, and c = -3, we can find the positive value of t.
Plugging in the values:
t = (-20 ± √(20^2 - 4(-4.9)(-3))) / (2(-4.9))
Simplifying further:
t = (-20 ± √(400 - 58.8)) / (-9.8)
t = (-20 ± √341.2) / (-9.8)
Calculating the square root:
t ≈ (-20 ± 18.47) / (-9.8)
We have two values for t:
t ≈ 2.217s and t ≈ 0.225s
Since the ball was initially hit upwards, we can discard the negative value of t.
Therefore, the ball has a displacement of 3m at approximately 2.217 seconds.
b) To calculate how long the ball will be above a height of 3m, we need to consider the time it takes for the ball to reach its maximum height and then descend back down.
The time it takes for the ball to reach its maximum height is given by:
v = u + at
Where:
v = final velocity (0m/s, at maximum height)
u = initial velocity (20m/s)
a = acceleration due to gravity (-9.8m/s^2)
Let's substitute the values into the equation:
0 = 20 + (-9.8)t
Rearranging the equation:
9.8t = 20
t = 20 / 9.8 ≈ 2.041s
So, it takes approximately 2.041 seconds for the ball to reach its maximum height.
To calculate the total time the ball will be above a height of 3m, we add the time it takes to reach the maximum height (2.041s) to the time it takes for the ball to fall from that height back to 3m.
Since the ball takes the same amount of time to descend from its maximum height as it took to reach it, the total time the ball will be above 3m is:
2.041s + 2.041s = 4.082s
Therefore, the ball will be above a height of 3m for approximately 4.082 seconds.
To solve this problem, we can use the equations of motion for vertically thrown objects.
a) To find the time when the ball has a displacement of 3m, we can use the equation:
s = ut + (1/2)at^2
Where:
- s is the displacement (3m in this case)
- u is the initial velocity (20m/s)
- a is the acceleration due to gravity (-9.8m/s^2, assuming we are near the Earth's surface)
- t is the time we need to find
Rearranging the equation to solve for t, we get:
t = (-u ± √(u^2 - 2as)) / a
Plugging in the given values, we have:
t = (-20 ± √(20^2 - 2*(-9.8)*3)) / (-9.8)
t = (-20 ± √(400 + 58.8)) / (-9.8)
t = (-20 ± √458.8) / (-9.8)
Now, there are two possible solutions due to the ± sign. Let's calculate both:
t₁ = (-20 + √458.8) / (-9.8)
t₂ = (-20 - √458.8) / (-9.8)
Calculating t₁:
t₁ = (-20 + √458.8) / (-9.8)
≈ 2.71 seconds
Calculating t₂:
t₂ = (-20 - √458.8) / (-9.8)
≈ -3.28 seconds
Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball will have a displacement of 3m at approximately 2.71 seconds.
b) To calculate the time the ball will be above a height of 3m, we need to consider the motion of the ball after reaching the peak of its trajectory. At that point, its velocity will be zero before it starts falling back towards the ground.
The time taken to reach the peak can be found using the equation:
v = u + at
Where:
- v is the final velocity (zero when reaching the peak)
- u is the initial velocity (20m/s)
- a is the acceleration due to gravity (-9.8m/s^2)
- t is the time taken to reach the peak
Setting v to zero:
0 = 20 - 9.8t
Solving for t:
t = 20 / 9.8
≈ 2.04 seconds
So, the ball takes approximately 2.04 seconds to reach the peak of its trajectory.
Now, to calculate the time the ball will be above a height of 3m, we need to consider the time it takes for the ball to fall back down from the peak to a height of 3m.
Using the equation:
s = ut + (1/2)at^2
Where:
- s is the displacement (-3m in this case because the ball is falling downwards)
- u is the initial velocity (zero when reaching the peak)
- a is the acceleration due to gravity (-9.8m/s^2)
- t is the time taken for the ball to fall from the peak to a height of 3m
Rearranging the equation to solve for t, we get:
t = √(2s/a)
Plugging in the values, we have:
t = √(2*(-3)/(-9.8))
≈ 0.78 seconds
Therefore, the ball will be above a height of 3m for approximately 0.78 seconds.
Note: The negative sign in the equation for time when the ball reaches a height of 3m is ignored because we're only interested in the time after the ball is hit upwards.