At the beginning of a basketball game, a referee tosses the ball straight up with a speed of 5.4 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?
(5.4 m/s)/g
That is how long it takes for the speed to reach zero, which occurs at the maximum height.
g is the acceleration of gravity. I assume you know its value.
0.447
To find the minimum time that a player must wait before touching the ball, we need to determine the time it takes for the ball to reach its maximum height and begin to fall down.
We can start by using the equation of motion for vertical motion:
V_f^2 = V_i^2 + 2aΔd
Where:
- V_f is the final velocity (which is 0 m/s at maximum height)
- V_i is the initial velocity (given as 5.4 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2)
- Δd is the change in position (height)
At maximum height, the final velocity is 0, so we have:
0 = (5.4 m/s)^2 + 2(-9.8 m/s^2)Δd
Simplifying the equation:
0 = 29.16 m^2/s^2 - 19.6 m/s^2 Δd
19.6 m/s^2 Δd = 29.16 m^2/s^2
Δd = 29.16 m^2/s^2 / 19.6 m/s^2
Δd = 1.49 m
This means the ball reaches a maximum height of 1.49 meters.
Next, we can use the equation of motion for vertical motion to find the time it takes for the ball to reach its maximum height:
V_f = V_i + at
Where:
- V_f is the final velocity (which is 0 m/s at maximum height)
- V_i is the initial velocity (given as 5.4 m/s)
- a is the acceleration due to gravity (-9.8 m/s^2)
- t is the time
0 = 5.4 m/s + (-9.8 m/s^2) t
Solving for t:
-5.4 m/s = -9.8 m/s^2 t
t = -5.4 m/s / -9.8 m/s^2
t ≈ 0.55 s
So, the ball takes approximately 0.55 seconds to reach its maximum height.
Finally, the minimum time that a player must wait before touching the ball is equal to the time it takes for the ball to reach its maximum height and start falling back down, which is approximately 0.55 seconds.