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 0.5 seconds after being thrown upward from the edge of a cliff with an initial velocity of 6 m/s, we can use the kinematic equation:

\[ v = u + at \]

where:
- \( v \) = final velocity (what we want to find)
- \( u \) = initial velocity (6 m/s, since the ball is thrown upward)
- \( a \) = acceleration (which is -g, since gravity acts downward; here \( g = 10 \, \text{m/s}^2 \), so \( a = -10 \, \text{m/s}^2 \))
- \( t \) = time (0.5 seconds)

Now, let's substitute the values into the equation step by step:

1. **Identify the initial velocity (u)**:
\[ u = 6 \, \text{m/s} \]

2. **Identify the acceleration (a)**:
\[ a = -10 \, \text{m/s}^2 \]

3. **Identify the time (t)**:
\[ t = 0.5 \, \text{s} \]

4. **Substitute the values into the kinematic equation**:
\[ v = u + at \]
\[ v = 6 \, \text{m/s} + (-10 \, \text{m/s}^2) \times (0.5 \, \text{s}) \]

5. **Calculate the acceleration term**:
\[ -10 \, \text{m/s}^2 \times 0.5 \, \text{s} = -5 \, \text{m/s} \]

6. **Now substitute back into the equation**:
\[ v = 6 \, \text{m/s} - 5 \, \text{m/s} \]

7. **Calculate the final velocity**:
\[ v = 1 \, \text{m/s} \]

Thus, the speed of the ball after 0.5 seconds is **1 m/s**.