A basketball player grabbing a rebound jumps 77 cm vertically.
(a) How much (total) time does the player spend in the top 22 cm of this jump?
(b) How much (total) time does the player spend in the bottom 22 cm of this jump?
v^2/2g = .77
v = 3.88 m/s
3.88t - 4.9t^2 > 55
0.185 < t < 0.607
(a) 0.607-0.185 = ___
(b) 0.792 - (a) = ___
To solve these questions, we need to use the principles of physics and motion equations.
Let's analyze each part of the jump separately:
(a) The time spent in the top 22 cm of the jump:
We can use the equation for vertical displacement in free-fall motion:
Δy = v₀t + (1/2)gt²
where:
Δy = vertical displacement
v₀ = initial velocity (0 m/s)
t = time
g = acceleration due to gravity (9.8 m/s²)
In this case, the vertical displacement Δy is 22 cm or 0.22 meters.
0.22 = 0 + (1/2) * 9.8 * t²
Rearranging the equation:
0.22 = 4.9t²
t² = 0.22 / 4.9
t ≈ √(0.045)
t ≈ 0.21 seconds
Therefore, the player spends approximately 0.21 seconds in the top 22 cm of the jump.
(b) The time spent in the bottom 22 cm of the jump:
We can use the same equation for vertical displacement as before:
Δy = v₀t + (1/2)gt²
In this case, the vertical displacement Δy is also 22 cm or 0.22 meters.
0.22 = 0 + (1/2) * 9.8 * t²
t² = 0.22 / 4.9
t ≈ √(0.045)
t ≈ 0.21 seconds
Therefore, the player spends approximately 0.21 seconds in the bottom 22 cm of the jump.
To summarize:
(a) The player spends approximately 0.21 seconds in the top 22 cm of the jump.
(b) The player also spends approximately 0.21 seconds in the bottom 22 cm of the jump.
To find the time spent in the top and bottom portions of the player's jump, we need to first determine the time it takes to reach the top and bottom heights of 22 cm. We can use the equation of motion for vertical motion to solve for time.
The equation of motion for vertical motion is:
h = ut + (1/2)gt^2
Where:
h is the height
u is the initial velocity
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
Let's calculate the time it takes for the player to reach the top and bottom heights.
(a) Time spent in the top 22 cm:
Initial height (h) = 0
Final height (h) = 22 cm = 0.22 m
Acceleration (g) = -9.8 m/s^2 (acceleration due to gravity)
Using the equation of motion, we can find the time (t) taken to reach the top height:
0.22 = 0 + (1/2)(-9.8)t^2
Simplifying the equation:
0.22 = -4.9t^2
Rearranging the equation:
t^2 = 0.22 / -4.9
Taking the square root on both sides to find t:
t = √(0.22 / -4.9)
Calculating the time t:
t ≈ 0.096 seconds (rounded to 3 decimal places)
Therefore, the basketball player spends approximately 0.096 seconds in the top 22 cm of the jump.
(b) Time spent in the bottom 22 cm:
To find the time spent in the bottom 22 cm, we can use the same approach as above.
Initial height (h) = 0.77 m
Final height (h) = 0.55 m (0.77 - 0.22 = 0.55)
Acceleration (g) = -9.8 m/s^2 (acceleration due to gravity)
Using the equation of motion, we can find the time (t) taken to reach the bottom height:
0.55 = 0.77 + (1/2)(-9.8)t^2
Simplifying the equation:
0.55 = 0.77 - 4.9t^2
Rearranging the equation:
t^2 = (0.77 - 0.55) / -4.9
Taking the square root on both sides to find t:
t = √((0.77 - 0.55) / -4.9)
Calculating the time t:
t ≈ 0.0925 seconds (rounded to 4 decimal places)
Therefore, the basketball player spends approximately 0.0925 seconds in the bottom 22 cm of the jump.