Ball Projection: Understanding Vertical Motion

When a ball is projected vertically upwards, it experiences acceleration due to gravity acting downwards. This causes its upward velocity to decrease until it momentarily stops at its maximum height. After reaching the maximum height, the ball starts falling back towards the ground, accelerating downwards due to gravity.

Velocity Analysis for a Vertically Projected Ball

The question states that a ball is projected up vertically with an initial velocity of \(9.8 \text{ m/s}\). We need to determine the total time it takes for this ball to reach the ground. To solve this, we can break the motion into two parts:

• Upward Journey: From the point of projection to the maximum height.
• Downward Journey: From the maximum height back to the ground.

We will use the standard value for the acceleration due to gravity, \(g = 9.8 \text{ m/s}^2\).

Time to Reach Maximum Height

First, let's calculate the time taken for the ball to reach its maximum height. At the maximum height, the final velocity of the ball becomes zero.

• Initial velocity, \(u = 9.8 \text{ m/s}\) (upwards)
• Final velocity at max height, \(v = 0 \text{ m/s}\)
• Acceleration due to gravity, \(a = -g = -9.8 \text{ m/s}^2\) (negative because it opposes the upward motion)

Using the first equation of motion:

\[v = u + at\]

Substituting the values:

\[0 = 9.8 + (-9.8)t_{up}\]

\[0 = 9.8 - 9.8t_{up}\]

Rearranging the equation to solve for \(t_{up}\) (time taken to go up):

\[9.8t_{up} = 9.8\]

\[t_{up} = \frac{9.8}{9.8}\]

\[t_{up} = 1 \text{ s}\]

So, the ball takes 1 second to reach its maximum height.

Time to Return to the Ground

For a vertically projected ball, neglecting air resistance, the motion is symmetrical. This means the time taken for the ball to go from the projection point to its maximum height is equal to the time it takes to fall back from the maximum height to the initial projection level.

Therefore, the time taken for the ball to fall from the maximum height back to the ground is:

\[t_{down} = t_{up} = 1 \text{ s}\]

Total Time to Reach the Ground

The total time the ball takes to reach the ground is the sum of the time taken for the upward journey and the downward journey.

\[T_{total} = t_{up} + t_{down}\]

\[T_{total} = 1 \text{ s} + 1 \text{ s}\]

\[T_{total} = 2 \text{ s}\]

Summary of Motion

Thus, the ball takes 2 seconds to reach the ground after being projected up vertically with an initial velocity of 9.8 m/s.