In uniformly accelerated motion the following equations 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 Its velocity afer 3 seconds ?

Options

A. 14.4 m/sec. downward
B. 14.4 m/sec. upward
C. 29.4 m/sec. downward
D. 29.4 m/sec. upward

Explanation

To solve the problem of finding the velocity of a ball projected upward after 3 seconds, we will use the equations of uniformly accelerated motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (Vo) = 15 m/s (upward)
- Acceleration (a) = -9.8 m/s² (downward, due to gravity)
- Time (t) = 3 seconds

### Step 2: Use the first equation of motion
The first equation of motion is given by:
\[ V = V_0 + at \]
Where:
- V = final velocity
- \( V_0 \) = initial velocity
- a = acceleration
- t = time

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ V = 15 \, \text{m/s} + (-9.8 \, \text{m/s}^2) \times (3 \, \text{s}) \]

### Step 4: Calculate the acceleration term
Calculate the term \( (-9.8 \, \text{m/s}^2) \times (3 \, \text{s}) \):
\[ -9.8 \times 3 = -29.4 \, \text{m/s} \]

### Step 5: Complete the calculation for V
Now substitute this value back into the equation:
\[ V = 15 \, \text{m/s} - 29.4 \, \text{m/s} \]
\[ V = 15 - 29.4 = -14.4 \, \text{m/s} \]

### Step 6: Interpret the result
The negative sign indicates that the velocity is directed downward. Therefore, the velocity of the ball after 3 seconds is:
\[ V = 14.4 \, \text{m/s} \, \text{(downward)} \]

### Final Answer
The velocity of the ball after 3 seconds is **14.4 m/s downward**.

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