An entertainer juggles balls while doing other activities. In one act, she throws a ball vertically upward, and while it is in the air, she runs to and from a table 5.20m away at a constant speed of 2.40m/s, returning just in time to catch the falling ball.
A) With what minimum initial speed must she throw the ball upward to accomplish this feat?
B) How high above its initial position is the ball just as she reaches the table?
18.87
To answer these two questions, we can use the principles of projectile motion. Let's break it down step by step.
A) With what minimum initial speed must she throw the ball upward to accomplish this feat?
In this scenario, the entertainer is throwing the ball vertically upward and then running to and from a table. To determine the minimum initial speed required, we need to consider the distance the entertainer has to cover and the time taken by the ball to reach its maximum height and fall back down.
1. First, let's find the time it takes for the ball to reach its maximum height. We can use the formula:
**t = (v_f - v_i) / g**
where:
- t is the time
- v_f is the final velocity (which is 0 when the ball reaches maximum height)
- v_i is the initial velocity
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the ball is thrown vertically upward, the final velocity will be 0 when it reaches its maximum height.
Therefore, t = (0 - v_i) / (-9.8) = v_i / 9.8
2. Once we have the time taken, we need to calculate the distance covered by the entertainer while the ball is in the air. The total distance covered is twice the distance between the table and the starting point, so it is 2 * 5.20m = 10.40m.
Since the entertainer covers this distance at a constant speed of 2.40m/s, we can calculate the time taken using the formula:
**t = d / v**
where:
- t is the time
- d is the distance
- v is the speed
Therefore, t = 10.40m / 2.40m/s = 4.33s (rounded to two decimal places)
3. The total time for the entertainer to catch the ball is the sum of the time taken for the ball to reach maximum height (from step 1) and the time taken for the entertainer to cover the table distance (from step 2). So, the total time is:
T = t + t = v_i / 9.8 + 4.33s
Now, to accomplish this feat, the entertainer must throw the ball upward with just enough initial speed to reach its maximum height within the given timeframe.
B) How high above its initial position is the ball just as she reaches the table?
Since the entertainer reaches the table just as the ball is about to fall back down, the height above its initial position will be the maximum height the ball reaches. To calculate this height, we can use the kinematic equation:
**h = v_i^2 / (2 * g)**
where:
- h is the height
- v_i is the initial velocity of the ball
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Now, we have the equation for the total time (from step 3):
T = v_i / 9.8 + 4.33s
We can substitute this expression for v_i in the equation for height:
h = (9.8 * (T - 4.33s))^2 / (2 * 9.8)
Simplifying the equation would give you the height above its initial position.
Please note that due to rounding and approximation, the final values may have slight variations.