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?
I don't understand this problem at all! Can someone help?
The max ht. occurs at V = 0.
V = Vo + gt = 0.
5.4 - 9.8t = 0
9.8t = 5.4
t = 0.55 s.
Sure! Let me help you understand this problem.
In this scenario, the referee tosses the ball straight up into the air with an initial speed of 5.4 m/s. As the ball moves upward, its speed decreases due to the force of gravity pulling it downward. Eventually, the ball reaches its maximum height and begins to fall back down.
To find the minimum time that a player must wait before touching the ball, we need to find two things: the time it takes for the ball to reach its maximum height, and the time it takes for the ball to fall back down to a level where the player can touch it.
First, let's find the time it takes for the ball to reach its maximum height. We can use the fact that the vertical motion of the ball is affected by gravity and is described by the kinematic equation:
vf = vi + at
Where:
vf = final velocity (which is 0 m/s at the maximum height)
vi = initial velocity (5.4 m/s)
a = acceleration (due to gravity, which is approximately -9.8 m/s^2)
Substituting these values into the equation, we can solve for the time it takes for the ball to reach its maximum height:
0 = 5.4 - 9.8t_max
Solving for t_max, we get:
t_max = 5.4 / 9.8
Next, we need to find the time it takes for the ball to fall back down to a level where the player can touch it. Since the ball was thrown straight up, it will take the same amount of time to fall back down as it took to reach its maximum height. So, the time it takes for the ball to fall back down is also t_max.
Therefore, the minimum time that a player must wait before touching the ball is t_max, which is approximately 0.55 seconds.
Sure! Let's break down the problem step by step.
1. The referee tosses the ball straight up with a speed of 5.4 m/s. This means that the initial velocity of the ball is 5.4 m/s.
2. Since the ball is thrown straight up, it will eventually reach its maximum height and start falling back down. At the highest point, the ball will momentarily come to a stop before changing direction.
3. We need to find the minimum time a player must wait before touching the ball. To do this, we need to determine how long it takes for the ball to reach its maximum height.
4. At the maximum height, the ball's vertical velocity becomes zero. Using the equation for average velocity, which states that average velocity is equal to the change in displacement divided by the change in time, we can find the time it takes for the ball to reach its maximum height.
5. In this case, since the ball's initial velocity is 5.4 m/s and its final velocity is 0 m/s, the average velocity would be half of the initial velocity.
Average velocity = (Initial velocity + Final velocity) / 2
0 m/s = (5.4 m/s + 0 m/s) / 2
Solving for time, we get:
(5.4 m/s + 0 m/s) / 2 = 2.7 m/s = 9.8 m/s^2 × time
time = (2.7 m/s) / (9.8 m/s^2) ≈ 0.275 seconds
Therefore, the time it takes for the ball to reach its maximum height is approximately 0.275 seconds.
6. Since the player cannot touch the ball until after it reaches its maximum height and starts falling back down, the player would need to wait at least 0.275 seconds before touching the ball.
So, the minimum time that a player must wait before touching the ball is approximately 0.275 seconds.