In uniformly accelerated motira the following questions hold :
V=Vo + at
`X=Vot + 1//2 at^2`
When X = displacement , V=Velocity at time, Vo= initial velocity, t=time , and a =acceleration. A ball is projected directly upward at a velocity of 15 m/sec.
What is the distance above the ground after 3 seconds ?

Options

A. 1.8 m
B. 0.9 m
C. 0 m
D. 3.6 m.

Explanation

To solve the problem of finding the distance above the ground after 3 seconds for a ball projected directly upward with an initial velocity of 15 m/s, we can use the second equation of motion:

\[ X = V_0 t + \frac{1}{2} a t^2 \]

Where:
- \( X \) = displacement
- \( V_0 \) = initial velocity
- \( t \) = time
- \( a \) = acceleration

### Step 1: Identify the values
- Initial velocity (\( V_0 \)) = 15 m/s (upward)
- Time (\( t \)) = 3 seconds
- Acceleration (\( a \)) = -10 m/s² (downward, due to gravity)

### Step 2: Substitute the values into the equation
Using the equation:

\[ X = V_0 t + \frac{1}{2} a t^2 \]

Substituting the known values:

\[ X = 15 \times 3 + \frac{1}{2} \times (-10) \times (3^2) \]

### Step 3: Calculate each term
1. Calculate \( 15 \times 3 \):
\[ 15 \times 3 = 45 \, \text{m} \]

2. Calculate \( \frac{1}{2} \times (-10) \times (3^2) \):
\[ 3^2 = 9 \]
\[ \frac{1}{2} \times (-10) \times 9 = -45 \, \text{m} \]

### Step 4: Combine the results
Now, combine the results from the two calculations:

\[ X = 45 - 45 \]

### Step 5: Final calculation
\[ X = 0 \, \text{m} \]

### Conclusion
The distance above the ground after 3 seconds is 0 meters, indicating that the ball has reached its maximum height and is returning to the ground.

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