Question
A student drops a ball from a height of 60 meters. The table below shows the ball’s speed and height as time passes.
Time (s):
0.0
0.5
1.0
1.5
2.0
2.5
Speed (m/s):
0
7
10
15
20
0
Height (m):
60
53
43
28
8
0
At what point in time does the ball have the highest kinetic energy?
A). At the start (0 seconds)
B). After 0.5 seconds
C). After 2 seconds
D). After 2.5 seconds
Time (s):
0.0
0.5
1.0
1.5
2.0
2.5
Speed (m/s):
0
7
10
15
20
0
Height (m):
60
53
43
28
8
0
At what point in time does the ball have the highest kinetic energy?
A). At the start (0 seconds)
B). After 0.5 seconds
C). After 2 seconds
D). After 2.5 seconds
Answers
GPT-4o mini
To determine at which point in time the ball has the highest kinetic energy, we need to consider the relationship between speed and kinetic energy. The formula for kinetic energy (\(KE\)) is:
\[
KE = \frac{1}{2} m v^2
\]
where \(m\) is the mass of the object and \(v\) is the velocity (speed) of the object. Since the mass of the ball remains constant throughout its fall, we only need to focus on the speed of the ball at each time point to determine the kinetic energy.
Let's summarize the speeds at the given time points:
- At 0.0 seconds: Speed = 0 m/s
- At 0.5 seconds: Speed = 7 m/s
- At 1.0 seconds: Speed = 10 m/s
- At 1.5 seconds: Speed = 15 m/s
- At 2.0 seconds: Speed = 20 m/s
- At 2.5 seconds: Speed = 0 m/s (the ball has reached the ground)
Now, we can compute the speeds at different time points to see where the speed is highest:
- At 0.0 seconds: \(v = 0\) m/s → \(KE = 0\)
- At 0.5 seconds: \(v = 7\) m/s → \(KE = \frac{1}{2} m (7^2) = \frac{49}{2} m\)
- At 1.0 seconds: \(v = 10\) m/s → \(KE = \frac{1}{2} m (10^2) = 50 m\)
- At 1.5 seconds: \(v = 15\) m/s → \(KE = \frac{1}{2} m (15^2) = \frac{225}{2} m\)
- At 2.0 seconds: \(v = 20\) m/s → \(KE = \frac{1}{2} m (20^2) = 200 m\)
- At 2.5 seconds: \(v = 0\) m/s → \(KE = 0\)
From the values calculated, we can see:
- At 0.5 seconds: \(KE = \frac{49}{2} m\)
- At 1.0 seconds: \(KE = 50m\)
- At 1.5 seconds: \(KE = \frac{225}{2} m\)
- At 2.0 seconds: \(KE = 200 m\)
- At 2.5 seconds: \(KE = 0\)
The maximum kinetic energy occurs at:
**2.0 seconds** with a speed of **20 m/s**.
Hence, the correct answer is:
**C). After 2 seconds**
\[
KE = \frac{1}{2} m v^2
\]
where \(m\) is the mass of the object and \(v\) is the velocity (speed) of the object. Since the mass of the ball remains constant throughout its fall, we only need to focus on the speed of the ball at each time point to determine the kinetic energy.
Let's summarize the speeds at the given time points:
- At 0.0 seconds: Speed = 0 m/s
- At 0.5 seconds: Speed = 7 m/s
- At 1.0 seconds: Speed = 10 m/s
- At 1.5 seconds: Speed = 15 m/s
- At 2.0 seconds: Speed = 20 m/s
- At 2.5 seconds: Speed = 0 m/s (the ball has reached the ground)
Now, we can compute the speeds at different time points to see where the speed is highest:
- At 0.0 seconds: \(v = 0\) m/s → \(KE = 0\)
- At 0.5 seconds: \(v = 7\) m/s → \(KE = \frac{1}{2} m (7^2) = \frac{49}{2} m\)
- At 1.0 seconds: \(v = 10\) m/s → \(KE = \frac{1}{2} m (10^2) = 50 m\)
- At 1.5 seconds: \(v = 15\) m/s → \(KE = \frac{1}{2} m (15^2) = \frac{225}{2} m\)
- At 2.0 seconds: \(v = 20\) m/s → \(KE = \frac{1}{2} m (20^2) = 200 m\)
- At 2.5 seconds: \(v = 0\) m/s → \(KE = 0\)
From the values calculated, we can see:
- At 0.5 seconds: \(KE = \frac{49}{2} m\)
- At 1.0 seconds: \(KE = 50m\)
- At 1.5 seconds: \(KE = \frac{225}{2} m\)
- At 2.0 seconds: \(KE = 200 m\)
- At 2.5 seconds: \(KE = 0\)
The maximum kinetic energy occurs at:
**2.0 seconds** with a speed of **20 m/s**.
Hence, the correct answer is:
**C). After 2 seconds**