A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend?
(a) In the top 15.0 cm of this jump and
(b) In the bottom 15.0cm? Do your results explain why such players seem to hang in the at at the top of a jump?
The Correct Answer and Explanation is :
To solve this problem, we analyze the motion of the basketball player under the laws of kinematics.
Known values:
- Maximum vertical height ((h_{\text{max}})) = 76.0 cm = 0.76 m
- Acceleration due to gravity ((g)) = (9.8 \, \text{m/s}^2)
The motion is symmetric, so the time spent during ascent and descent is the same. The total time ((T_{\text{total}})) will be calculated, followed by times for specific height ranges.
(a) Time in the top 15.0 cm of the jump
The top 15.0 cm corresponds to the height range from (h = 0.61 \, \text{m}) to (h_{\text{max}} = 0.76 \, \text{m}).
Velocity at (h_{\text{max}}):
At maximum height, velocity is (v = 0).
Velocity at (h = 0.61 \, \text{m}):
From energy conservation:
[
v^2 = 2g(h_{\text{max}} – h)
]
[
v = \sqrt{2 \cdot 9.8 \cdot (0.76 – 0.61)} = \sqrt{2 \cdot 9.8 \cdot 0.15} \approx 1.71 \, \text{m/s}
]
Time to travel through the top 15.0 cm:
The time for this segment ((t_{\text{top}})) is calculated using:
[
t = \frac{v}{g}
]
[
t_{\text{top}} = \frac{1.71}{9.8} \approx 0.174 \, \text{s}
]
Since this is the time for ascent or descent, the total time in the top 15.0 cm is:
[
T_{\text{top}} = 2 \cdot t_{\text{top}} \approx 0.348 \, \text{s}
]
(b) Time in the bottom 15.0 cm of the jump
The bottom 15.0 cm corresponds to the height range from (h = 0) to (h = 0.15 \, \text{m}).
Velocity at (h = 0.15 \, \text{m}):
[
v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 0.15} \approx 1.71 \, \text{m/s}
]
Velocity at (h = 0) (takeoff/landing):
[
v = 0
]
Time for this segment ((t_{\text{bottom}})):
[
t_{\text{bottom}} = \frac{1.71}{9.8} \approx 0.174 \, \text{s}
]
For both ascent and descent:
[
T_{\text{bottom}} = 2 \cdot t_{\text{bottom}} \approx 0.348 \, \text{s}
]
Analysis and Explanation
The time spent in the top 15.0 cm ((T_{\text{top}} \approx 0.348 \, \text{s})) is equal to the time spent in the bottom 15.0 cm ((T_{\text{bottom}} \approx 0.348 \, \text{s})). However, players seem to “hang” in the air because their vertical speed near the top is much slower than near the bottom. This creates the illusion of prolonged suspension. At the bottom, their speed changes rapidly due to higher acceleration, making this phase seem quicker.