In uniformly accelerated motion 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 highest point this ball will reach?

Options

A. a. 38.66 m
B. b.11.48 m
C. c. 9.80 m
D. d. 1.53 m.

Explanation

To find the highest point a ball reaches when projected directly upward with an initial velocity of 15 m/s, we can use the equations of uniformly accelerated motion. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Initial velocity (Vo) = 15 m/s (upward)
- Final velocity (V) at the highest point = 0 m/s (the ball stops momentarily at the highest point)
- Acceleration (a) = -g = -9.8 m/s² (the acceleration due to gravity acts downward)

### Step 2: Use the first equation of motion
We can use the equation:
\[ V = Vo + at \]
At the highest point, V = 0, so we can rearrange the equation to find the time (t) it takes to reach the highest point:
\[ 0 = 15 + (-9.8)t \]
\[ 9.8t = 15 \]
\[ t = \frac{15}{9.8} \]
\[ t \approx 1.53 \, \text{s} \]

### Step 3: Use the second equation of motion to find the maximum height (X)
Now, we can use the second equation of motion:
\[ X = Vo t + \frac{1}{2} a t^2 \]
Substituting the known values:
\[ X = 15 \cdot t + \frac{1}{2} \cdot (-9.8) \cdot t^2 \]
Substituting \( t \approx 1.53 \):
\[ X = 15 \cdot (1.53) + \frac{1}{2} \cdot (-9.8) \cdot (1.53)^2 \]

### Step 4: Calculate each term
1. Calculate \( 15 \cdot 1.53 \):
\[ 15 \cdot 1.53 \approx 22.95 \, \text{m} \]

2. Calculate \( \frac{1}{2} \cdot (-9.8) \cdot (1.53)^2 \):
\[ (1.53)^2 \approx 2.3409 \]
\[ \frac{1}{2} \cdot (-9.8) \cdot 2.3409 \approx -11.48 \, \text{m} \]

### Step 5: Combine the results
Now, substitute these values back into the equation for X:
\[ X \approx 22.95 - 11.48 \]
\[ X \approx 11.47 \, \text{m} \]

### Final Answer
The highest point the ball will reach is approximately **11.47 meters**.

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