A ball is thrown upward from edge of a cliff with an intial velocity of 6 m/s. How fast is it moving 1/2 s later ?`(g=10 m/s^(2))`

Answer

To solve the problem of how fast the ball is moving after 1/2 second when thrown upward with an initial velocity of 6 m/s, we can follow these steps:

### Step 1: Identify the given values
- Initial velocity (u) = 6 m/s (upward)
- Acceleration (a) = -g = -10 m/s² (since gravity acts downward)
- Time (t) = 1/2 s

### Step 2: Use the kinematic equation
We can use the kinematic equation that relates initial velocity, final velocity, acceleration, and time:
\[ v = u + at \]
where:
- \( v \) = final velocity
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

### Step 3: Substitute the known values into the equation
Substituting the values into the equation:
\[ v = 6 \, \text{m/s} + (-10 \, \text{m/s}^2) \cdot (0.5 \, \text{s}) \]

### Step 4: Calculate the acceleration component
Calculate \( at \):
\[ at = -10 \, \text{m/s}^2 \cdot 0.5 \, \text{s} = -5 \, \text{m/s} \]

### Step 5: Calculate the final velocity
Now substitute back into the equation for \( v \):
\[ v = 6 \, \text{m/s} - 5 \, \text{m/s} \]
\[ v = 1 \, \text{m/s} \]

### Step 6: Determine the direction of the final velocity
Since the final velocity is positive, the ball is still moving upward, but at a reduced speed of 1 m/s.

### Final Answer
The speed of the ball after 1/2 second is **1 m/s** upward.

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