Askiitians Tutor Team
Last Activity: 5 Months ago
Let's break down these two concepts related to projectile motion, specifically focusing on the behavior of an object projected vertically. We'll analyze the time taken to reach maximum height and the time taken for the ascent and descent of a ball thrown upward.
Time Taken to Reach Maximum Height
When a body is projected upward, it travels against the force of gravity until it reaches its maximum height. The key to understanding the time taken to travel the first half of the distance compared to the second half lies in the principles of motion under constant acceleration.
Understanding the Motion
Consider a body projected vertically with an initial velocity \( u \). The acceleration due to gravity \( g \) acts downward, opposing the motion. The equations of motion tell us that:
- The time taken to reach maximum height \( T \) can be calculated using the formula: \( T = \frac{u}{g} \).
- The maximum height \( H \) reached can be calculated using: \( H = \frac{u^2}{2g} \).
Dividing the Journey
Now, let's analyze the journey in two halves:
- The first half of the journey is from the ground to the height \( \frac{H}{2} \).
- The second half is from \( \frac{H}{2} \) to the maximum height \( H \).
Using the equations of motion, we can derive the time taken to reach \( \frac{H}{2} \). The time taken to reach any height \( h \) can be expressed as:
Using \( h = ut - \frac{1}{2}gt^2 \), we can set \( h = \frac{H}{2} \) and solve for time \( t_1 \) to reach \( \frac{H}{2} \) and \( t_2 \) to reach \( H \). The key insight is that the object is decelerating due to gravity, meaning it takes longer to cover the second half of the height.
Conclusion on Time Comparison
Mathematically, it can be shown that:
Time for the second half \( t_2 \) is greater than the time for the first half \( t_1 \). This is because as the object rises, it slows down, taking more time to cover the same distance in the second half of its journey.
Time of Ascent vs. Descent
Now, let's consider the second part of your question regarding the time taken for a ball thrown vertically upward. We denote the time taken to go up as \( T_1 \) and the time taken to come down as \( T_2 \).
Analyzing the Ascent and Descent
When the ball is thrown upward, it will rise until its velocity becomes zero at the maximum height. The time taken to reach this point is \( T_1 \). The ball then falls back down under the influence of gravity, accelerating downwards. The time taken to fall back to the ground from the maximum height is \( T_2 \).
Why \( T_1 < T_2 \)?
To understand why \( T_1 \) is less than \( T_2 \), consider the following:
- During ascent, the ball is moving against gravity, which decelerates it until it stops at the peak.
- During descent, the ball accelerates due to gravity, meaning it gains speed as it falls back down.
Using the equations of motion, we can see that the time taken to fall from a height \( H \) (which is the same height it ascended) is greater than the time taken to rise to that height. This is because the ball accelerates during its descent, covering the distance more quickly than it did while ascending.
Final Thoughts
In summary, both scenarios illustrate fundamental principles of kinematics. The time taken to travel the second half of the ascent is greater than the first half due to the deceleration caused by gravity. Similarly, the time taken to descend from maximum height is greater than the time taken to ascend, as the object accelerates downward. These principles are crucial in understanding motion under gravity.
