Calculating Time to Maximum Height in Vertical Motion

This question asks us to determine the time it takes for a ball, thrown vertically upward with a specific initial speed, to reach its maximum height. This scenario is a classic example of motion under constant acceleration due to gravity.

Understanding Vertical Projectile Motion

When an object is thrown vertically upward, it experiences a downward acceleration due to gravity. This acceleration causes the object's upward velocity to decrease continuously until it momentarily becomes zero at the highest point of its trajectory. After reaching the maximum height, the object starts falling back down.

Key points for motion to maximum height:

• The initial velocity is upward.
• The acceleration is downward (due to gravity), taken as negative if upward is positive.
• At the maximum height, the instantaneous velocity of the object is zero.

Identifying Given Information and Goal

• Initial velocity of the ball (\(u\)) = 40 m/s (upward)
• Acceleration due to gravity (\(g\)) = 9.8 m/s² (downward)
• Final velocity at maximum height (\(v\)) = 0 m/s
• We need to find the time taken (\(t\)) to reach this maximum height.

Since we are considering motion under constant acceleration, we can use the equations of kinematics. We need an equation that relates initial velocity, final velocity, acceleration, and time.

Applying the Correct Kinematic Equation

The appropriate kinematic equation is:

\(v = u + at\)

Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.

In our case, the acceleration \(a\) is the acceleration due to gravity, \(g\). Since the initial velocity is upward and gravity acts downward, we take \(u\) as positive and \(a = -g\).

So the equation becomes:

\(v = u - gt\)

Step-by-Step Calculation

We are given:

• \(u = 40\) m/s
• \(v = 0\) m/s (at maximum height)
• \(g = 9.8\) m/s²

Substitute these values into the equation \(v = u - gt\):

\(0 = 40 - (9.8) \times t\)

Now, we solve for \(t\):

\(9.8t = 40\)

\(t = \frac{40}{9.8}\)

Calculating the value of \(t\):

\(t \approx 4.0816\) seconds

Comparing with the Options

The calculated time is approximately 4.08 seconds. Let's look at the given options:

• 2 s
• 3 s
• 4 s
• 5 s

The value 4.08 seconds is closest to 4 seconds among the given options.

Conclusion on Time to Reach Maximum Height

Using the kinematic equation \(v = u + at\), and considering the final velocity is zero at the maximum height, we calculated the time taken for the ball to reach the maximum height to be approximately 4.08 seconds. Comparing this value with the provided options, 4 seconds is the closest approximation.

Revision Table: Key Concepts

Additional Information: Motion Symmetries and Gravity

Understanding vertical motion under gravity involves a few important points:

• Constant Acceleration: The acceleration due to gravity is considered constant near the Earth's surface (ignoring air resistance). This makes the kinematic equations applicable.
• Direction Matters: It is crucial to define a positive direction (e.g., upward) and consistently use signs for velocity, displacement, and acceleration. Since gravity acts downward, its acceleration is negative if upward is positive.
• Time Symmetry: In the absence of air resistance, the time taken for the object to reach its maximum height from the point of projection is equal to the time taken for it to fall back from the maximum height to the point of projection.
• Speed Symmetry: The speed of the object at any given height above the point of projection while going up is equal to its speed at the same height while coming down.

These principles help analyze the entire trajectory of a vertically projected object.