Baseball Throw From Cliff Problem Solving Maximum Height And Time
Introduction
In this article, we will analyze the scenario where Jayani throws a baseball straight up from a 15-meter high cliff with an initial velocity of 12 m/s. We will delve into the physics principles governing projectile motion to determine the maximum height the ball reaches relative to the ground and the total time the ball spends in the air. Understanding these concepts requires applying the equations of motion under constant acceleration, which in this case, is the acceleration due to gravity. This analysis not only helps in solving the problem at hand but also provides a broader understanding of how objects move under the influence of gravity.
Problem Statement
Jayani is standing on a cliff that is 15 meters high. She throws a baseball straight up into the air with an initial velocity of 12 meters per second. Our task is to solve two key aspects of this problem:
- Determine the maximum height the ball reaches with respect to the ground below the cliff.
- Calculate the total time the ball is in the air before it hits the ground.
1. Determining the Maximum Height of the Ball
To determine the maximum height, we need to consider the ball's motion from the moment it leaves Jayani's hand until it reaches its highest point. At the maximum height, the ball's velocity will momentarily be zero before it starts to descend. We can use the following kinematic equation, which relates final velocity (v), initial velocity (u), acceleration (a), and displacement (s):
v^2 = u^2 + 2as
Here, v (final velocity) is 0 m/s at the maximum height, u (initial velocity) is 12 m/s, a (acceleration) is the acceleration due to gravity, which is -9.8 m/s² (negative because it acts downwards), and s is the displacement we want to find. Plugging in the values:
0 = (12 m/s)^2 + 2(-9.8 m/s^2)s
Solving for s:
0 = 144 m^2/s^2 - 19.6 m/s^2 * s
19.6 m/s^2 * s = 144 m^2/s^2
s = 144 m^2/s^2 / 19.6 m/s^2
s ≈ 7.35 meters
This displacement s represents the height the ball rises above the point where Jayani threw it. Since Jayani is standing on a 15-meter cliff, we need to add this height to the cliff's height to find the maximum height with respect to the ground:
Maximum height above the ground = Cliff height + s
Maximum height above the ground = 15 meters + 7.35 meters
Maximum height above the ground ≈ 22.35 meters
Therefore, the baseball reaches a maximum height of approximately 22.35 meters above the ground.
2. Calculating the Total Time the Ball is in the Air
To calculate the total time the ball is in the air, we need to consider two phases of its motion: the time it takes to reach the maximum height and the time it takes to fall from the maximum height to the ground. We can use another kinematic equation:
v = u + at
First, let's find the time (t₁) it takes for the ball to reach its maximum height. At the maximum height, the final velocity v is 0 m/s, the initial velocity u is 12 m/s, and the acceleration a is -9.8 m/s²:
0 = 12 m/s + (-9.8 m/s^2) * t₁
Solving for t₁:
9.8 m/s^2 * t₁ = 12 m/s
t₁ = 12 m/s / 9.8 m/s^2
t₁ ≈ 1.22 seconds
So, it takes approximately 1.22 seconds for the ball to reach its maximum height. Now, we need to find the time (t₂) it takes for the ball to fall from the maximum height (22.35 meters) to the ground. For this part of the motion, we consider the initial velocity to be 0 m/s (since the ball momentarily stops at its peak), and the acceleration is 9.8 m/s² (positive because we are considering the downward motion). We use the following kinematic equation:
s = ut + (1/2)at^2
Where s is the displacement (22.35 meters), u is the initial velocity (0 m/s), a is the acceleration (9.8 m/s²), and t is the time we want to find. Plugging in the values:
22.35 m = (0 m/s) * t₂ + (1/2)(9.8 m/s^2) * t₂^2
22.35 m = 4.9 m/s^2 * t₂^2
Solving for t₂:
t₂^2 = 22.35 m / 4.9 m/s^2
t₂^2 ≈ 4.56 s^2
t₂ ≈ √4.56 s^2
t₂ ≈ 2.14 seconds
Thus, it takes approximately 2.14 seconds for the ball to fall from its maximum height to the ground. To find the total time the ball is in the air, we add the time it takes to go up (t₁) and the time it takes to come down (t₂):
Total time = t₁ + t₂
Total time ≈ 1.22 seconds + 2.14 seconds
Total time ≈ 3.36 seconds
Therefore, the total time the baseball is in the air is approximately 3.36 seconds.
Conclusion
In conclusion, by applying the principles of projectile motion and the kinematic equations, we have successfully determined that the baseball thrown by Jayani reaches a maximum height of approximately 22.35 meters above the ground and remains in the air for about 3.36 seconds. This problem illustrates the fundamental concepts of physics, including motion under constant acceleration and the effects of gravity on moving objects. Understanding these principles is crucial for solving a wide range of physics problems and for gaining insights into the world around us. The detailed step-by-step solution provided here offers a clear methodology for tackling similar problems in mechanics and highlights the practical application of physics in everyday scenarios.
