Understanding the Physics Problem

This problem asks us to find the magnitude of the displacement of a ball thrown vertically upward after a specific time (7 seconds). We are given the maximum height the ball reaches (80 m) and the value of gravitational acceleration (\(g = 10 \text{ m/s}^2\)). Displacement is the change in position from the starting point, including direction. Its magnitude is the straight-line distance between the initial and final positions.

Key Kinematics Concepts

To solve this, we will use the equations of motion under constant acceleration (due to gravity). Key concepts include:

• Displacement: The vector quantity representing the change in position. Unlike distance, it depends only on the initial and final points.
• Vertical Motion: Motion under the influence of gravity, where acceleration is constant (\(-g\)) when taking upward as positive.
• Maximum Height: The point where the vertical velocity momentarily becomes zero before the object starts falling down.
• Time of Flight: The total time an object spends in the air.

Step-by-Step Calculation

Step 1: Find the initial velocity (\(u\))

We know the ball reaches a maximum height (\(h\)) of 80 m. At the maximum height, the final velocity (\(v\)) is 0. Using the equation \(v^2 = u^2 + 2as\), where \(a = -g = -10 \text{ m/s}^2\) (taking upward as positive) and \(s = h = 80\) m:

The initial upward velocity is 40 m/s.

Step 2: Calculate the time taken to reach maximum height (\(t_{up}\))

Using the equation \(v = u + at\), where \(v=0\), \(u=40\) m/s, and \(a=-10\) m/s²:

It takes 4 seconds for the ball to reach its maximum height.

Step 3: Determine the total time of flight (\(T\))

For motion symmetrical about the peak (neglecting air resistance), the time taken to fall back to the original position is equal to the time taken to go up. So, \(t_{down} = t_{up} = 4\) s.

The total time of flight for the ball to return to its original position is 8 seconds.

Step 4: Analyze the position at \(t = 7\) seconds

The time we are interested in is 7 seconds. This is less than the total time of flight (8 seconds). This means the ball is still in the air at 7 seconds and has already passed its maximum height (at t=4s) and is now falling back down.

Step 5: Calculate the displacement at \(t = 7\) seconds

We can find the displacement at \(t=7\) s using the equation \(s = ut + \frac{1}{2}at^2\), where \(u=40\) m/s, \(a=-10\) m/s², and \(t=7\) s:

The displacement after 7 seconds is +35 m. The positive sign indicates that the final position is 35 m above the initial position (since we took upward as positive).

The question asks for the magnitude of the displacement.

Magnitude of displacement = \(|s(7)| = |35 \text{ m}| = 35 \text{ m}\).

Result

After 7 seconds of motion, the ball is 35 meters above its original starting position. The magnitude of its displacement is 35 m.

Revision Table: Key Values

Additional Information: Displacement vs. Distance

It's important to distinguish between displacement and distance traveled.

• Displacement: Only cares about the start and end points. In this case, after 7 seconds, the ball is 35 m above the start. The displacement vector points upward, and its magnitude is 35 m.
• Distance Traveled: The total path length covered. At 7 seconds, the ball went up 80 m (in 4s) and then fell down for 3 seconds (from t=4s to t=7s). The distance fallen in 3 seconds from rest (at peak) is \(s_{fall} = 0 \times 3 + \frac{1}{2}(10)(3)^2 = 5 \times 9 = 45\) m. The total distance traveled is \(80 \text{ m} + 45 \text{ m} = 125 \text{ m}\).

The question specifically asks for the magnitude of the displacement, which is 35 m.