Understanding Vertical Motion and Time to Highest Point

This problem asks us to find the time it takes for a ball, thrown vertically upward from the ground with a specific initial speed, to reach the highest point of its trajectory. This involves understanding the principles of vertical motion under constant acceleration due to gravity.

Problem Analysis: Ball Thrown Vertically Upward

When an object is thrown vertically upward, its initial velocity is directed upwards. As it travels upwards, the force of gravity acts downwards, causing the object to decelerate. This means its upward velocity decreases with time. At the highest point of its journey, the ball momentarily stops before starting to fall back down. At this exact moment at the peak, the ball's instantaneous velocity is zero.

Given Information

Let's list the information provided in the problem:

• Initial velocity of the ball (\(u\)) = 25.2 m/s (upwards)
• Acceleration due to gravity (\(g\)) = -9.8 m/s² (downwards, opposing upward motion)
• Final velocity of the ball at the highest point (\(v\)) = 0 m/s

We need to find the time (\(t\)) taken to reach this highest point.

Physics Principle: Kinematic Equations

We can use the kinematic equations, which describe the motion of objects under constant acceleration. The most suitable equation here, as it relates initial velocity, final velocity, acceleration, and time, is:

\(v = u + at\)

In vertical motion, the acceleration \(a\) is the acceleration due to gravity, \(g\). Since we define upward motion as positive and gravity acts downwards, \(g\) is taken as negative (\(-9.8 \, \text{m/s}^2\)) when the initial velocity is positive.

So the equation becomes:

\(v = u + gt\)

Step-by-Step Solution for Time to Highest Point

We can now substitute the known values into the equation \(v = u + gt\) and solve for \(t\).

• Set the final velocity \(v\) to 0 m/s (velocity at the highest point).
• Substitute the initial velocity \(u = 25.2 \, \text{m/s}\).
• Substitute the acceleration due to gravity \(g = -9.8 \, \text{m/s}^2\).
• Solve the resulting equation for \(t\).

Calculation of Time

Using the equation \(v = u + gt\):

\(0 = 25.2 \, \text{m/s} + (-9.8 \, \text{m/s}^2)t\)

Rearrange the equation to solve for \(t\):

\(0 = 25.2 - 9.8t\)

\(9.8t = 25.2\)

\(t = \frac{25.2}{9.8}\)

Calculating the value:

\(t \approx 2.5714 \, \text{s}\)

Result and Conclusion

The time taken for the ball to reach the highest point is approximately 2.57 seconds.

Comparing this value with the given options, we find that it matches one of them.

Let's summarise the key values:

Using the formula \(v = u + gt\), we calculated \(t = 2.57 \, \text{s}\).

Revision Table: Key Concepts for Vertical Motion

Additional Information: Factors Affecting Vertical Throw

While this problem simplifies the scenario, in reality, factors like air resistance can affect the motion of the ball. Air resistance opposes the motion, reducing the speed more quickly, and would mean the time to reach the highest point would be slightly different from the value calculated using only gravity.

Also, the acceleration due to gravity \(g\) is approximately 9.8 m/s², but it can vary slightly depending on location and altitude. For most standard physics problems at ground level, 9.8 m/s² is a commonly used value.

Understanding the concept that velocity is zero at the highest point is crucial for solving many vertical motion problems.