A body is rpojected vertically upwards with a velocity of `10m//s`. If reaches the maximum height (h) in time (t). In time ` t//2` the height covered is (g= 10 m//s^@)`

Options

A. (a) ` h//2`
B. (b) (2 //5) h`
C. (c ) ` (3 //4) h`
D. (d) ` ( 5 //8) h`

Explanation

To solve the problem step by step, we need to analyze the motion of the body projected vertically upwards with an initial velocity of 10 m/s. We will find the maximum height reached and the height covered in half the time taken to reach that maximum height.

### Step 1: Determine the time taken to reach maximum height (t)
The body is projected upwards with an initial velocity (u) of 10 m/s. At maximum height, the final velocity (v) becomes 0 m/s. We can use the first equation of motion:

\[ v = u + at \]

Where:
- \( v = 0 \) m/s (at maximum height)
- \( u = 10 \) m/s (initial velocity)
- \( a = -g = -10 \) m/s² (acceleration due to gravity)

Substituting the values:

\[ 0 = 10 - 10t \]

Rearranging gives:

\[ 10t = 10 \implies t = 1 \text{ second} \]

### Step 2: Calculate the maximum height (h)
Now, we can use the second equation of motion to find the maximum height (h):

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s = h \) (maximum height)
- \( u = 10 \) m/s
- \( a = -10 \) m/s²
- \( t = 1 \) s

Substituting the values:

\[ h = 10(1) + \frac{1}{2}(-10)(1^2) \]
\[ h = 10 - 5 = 5 \text{ meters} \]

### Step 3: Calculate the height covered in time \( t/2 \)
Now, we need to find the height covered in half the time, which is \( t/2 = 0.5 \) seconds. We will use the same equation of motion:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( t = 0.5 \) seconds

Substituting the values:

\[ s = 10(0.5) + \frac{1}{2}(-10)(0.5^2) \]
\[ s = 5 - \frac{1}{2}(10)(0.25) \]
\[ s = 5 - 1.25 = 3.75 \text{ meters} \]

### Step 4: Express height in terms of h
Since we found that the maximum height \( h = 5 \) meters, we can express the height covered in \( t/2 \) in terms of \( h \):

\[ s = 3.75 = \frac{3}{4} \times 5 = \frac{3h}{4} \]

### Final Answer
The height covered in time \( t/2 \) is \( \frac{3h}{4} \).

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