Displacement of a Ball Thrown Vertically: A Detailed Analysis

When a ball is thrown vertically into the air with an initial velocity of 10m/s, it undergoes a complex motion influenced by gravity. Understanding the displacement during the initial upward motion, the peak height, and the subsequent downward motion is crucial for grasping the dynamics of such motions.

Initial Motion and Maximum Height

Let's consider the ball thrown vertically with an initial velocity of 10m/s. The acceleration due to gravity, denoted as g -9.80665 m/s^2, acts downwards. At the start of the motion, the ball will decelerate under the influence of gravity.

The ball reaches its maximum height when its velocity becomes zero. This can be calculated using the following kinematic equation:

v u at

Where v is the final velocity (0 m/s at the peak height), u is the initial velocity (10 m/s), a is the acceleration due to gravity (-9.80665 m/s2), and t is the time.

At the peak height:

0 10 - 9.80665t

Solving for t:

t 1.019716 s

Plugging this value into the equation for the maximum height s:

s ut 0.5gt^2

s 10(1.019716) 0.5(-9.80665)(1.019716)^2

s 5.098581 m

Downward Motion and Displacement

After reaching the maximum height, the ball begins to fall under the constant acceleration due to gravity. The time for the downward motion is 1.5 seconds minus the time it took to reach the peak height:

t_{down} 1.5 - 1.019716 0.480283 s

During this interval, the ball will continue to fall. We can use the equation for displacement under constant acceleration:

s ut 0.5gt^2

Here, the initial velocity u is 0 m/s (since the ball starts from rest at the peak height), and the acceleration g is -9.80665 m/s2, which causes the ball to fall.

Therefore:

s_{down} 0(0.480283) 0.5(-9.80665)(0.480283)^2

s_{down} 1.131060 m

Note that this displacement s_{down} is the vertical distance the ball has fallen by the end of 0.480283 seconds.

Total Displacement

The total displacement of the ball from its initial height to its position 1.5 seconds after being thrown is the difference between the peak height and the distance fallen during the downward motion.

Total displacement:

s_{total} 5.098581 - 1.131060 3.967521 m

This displacement is upwards, indicating that the ball has not yet returned to its original position and has travelled 3.967521 meters above the initial point of release after 1.5 seconds.

Conclusion

Understanding the dynamics of ball motion involves grasping the concepts of initial velocity, deceleration, time of flight, maximum height, and displacement to accurately predict the ball's position at any given time. The analysis presented here demonstrates the application of basic kinematic equations to solve real-world problems in physics, particularly when dealing with vertical motion under the influence of gravity.