A basketball player standing near the basket to grap a rebound, jumps 62 cm vertically. How much total time does the player s in the top 15 cm of the jump? How much total time does the player spend in the bottm 15 cm of this jump?
If he jumps up H = 0.62 m, the initial velocity V is sucn that
(1/2) M V^2 = M g H
V = sqrt (2gH) = 3.48 m/s
Height vs time is given by
y = 3.38 t - 4.90 t^2
Solve for the times when H = 0.15 m and y = 0.47 m. You will need to use the postive root of the quadratic eauation.
At the top of the trajectory, H = 0.62 m and g t = V = 3.48 m/s, so t = 0.355 s there.
t = 0 when y = 0. Knowing the times when H = 0.15 and 0.47 m will let you calculate the intervals they are asking for.
To find the total time the player spends in the top and bottom 15 cm of the jump, we need to consider the player's vertical motion.
Let's break down the problem step by step:
1. First, let's determine the total time spent in the entire jump. We can use the formula for the time taken in vertical motion:
t = √(2h/g)
Where,
t is the time taken,
h is the height of the vertical jump, which is 62 cm (0.62 m) in this case,
and g is the acceleration due to gravity, approximately 9.8 m/s².
Substituting the values into the formula, we get:
t = √(2 * 0.62 / 9.8)
t ≈ 0.35 seconds (rounded to 2 decimal places)
2. The next step is to calculate the time spent in the top 15 cm of the jump. Since the player is moving upwards, this will occur during the first half of the total time. We can use the following equation:
time_top = (15 cm / 62 cm) * t
Where,
time_top is the time spent in the top 15 cm,
and t is the total time calculated earlier.
Substituting the values, we have:
time_top = (0.15 m / 0.62 m) * 0.35 s
time_top ≈ 0.08 seconds (rounded to 2 decimal places)
3. Similarly, we can calculate the time spent in the bottom 15 cm of the jump. This will occur during the second half of the total time. Using the same equation as before:
time_bottom = (15 cm / 62 cm) * t
Substituting the values, we get:
time_bottom = (0.15 m / 0.62 m) * 0.35 s
time_bottom ≈ 0.08 seconds (rounded to 2 decimal places)
Therefore, the basketball player spends approximately 0.08 seconds in both the top and bottom 15 cm of the jump.