A ball thrown vertically upwards after reaching a maximum height h returns to the starting point after a time of 10 s. Its displacement after 5 s is

Options

A. h
B. 2h
C. 10h
D. 20h

Explanation

To solve the problem, we need to analyze the motion of the ball thrown vertically upwards. Here’s a step-by-step solution:

### Step 1: Understand the Total Time of Flight
The total time of flight for the ball is given as 10 seconds. This means that it takes 10 seconds for the ball to go up to its maximum height and then return back to the starting point.

### Step 2: Determine the Time to Reach Maximum Height
Since the time taken to go up is equal to the time taken to come down, the time to reach the maximum height (h) is half of the total time of flight. Thus:
\[ t_{up} = \frac{10 \, \text{s}}{2} = 5 \, \text{s} \]

### Step 3: Calculate Displacement After 5 Seconds
Displacement is defined as the change in position. After 5 seconds, the ball has reached its maximum height and is at rest momentarily before it starts descending. At this point, the displacement is equal to the maximum height (h) reached by the ball.

### Step 4: Use the Kinematic Equation
We can use the kinematic equation for vertical motion:
\[ h = v_0 t - \frac{1}{2} g t^2 \]
Where:
- \(h\) is the maximum height,
- \(v_0\) is the initial velocity,
- \(g\) is the acceleration due to gravity (approximately \(9.81 \, \text{m/s}^2\)),
- \(t\) is the time taken to reach maximum height.

Since we are interested in the displacement after 5 seconds, we can conclude:
- At 5 seconds, the ball reaches its maximum height (h).

### Step 5: Conclusion
Thus, the displacement of the ball after 5 seconds is equal to the maximum height (h) it reaches.

\[ \text{Displacement after 5 seconds} = h \]