A body is projected vertically upwards with a velocity `u` it crosses a point in its journey at a height `h` twice just after 1 and 7 seconds. The value of `u` in `ms^(-1)` is `(g=10ms^(-2))`

Options

A. 50
B. 40
C. 30
D. 20

Explanation

To solve the problem step by step, we will use the equations of motion for a body projected vertically upwards. The key points are that the body crosses the same height \( h \) at two different times, \( t_1 = 1 \) second and \( t_2 = 7 \) seconds.

### Step 1: Write the equation of motion
The equation of motion for displacement \( s \) is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (height \( h \) in this case),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (which is \( -g \) for upward motion, where \( g = 10 \, \text{m/s}^2 \)),
- \( t \) is the time.

### Step 2: Set up the equations for both times
For \( t_1 = 1 \) second:
\[ h = u(1) + \frac{1}{2}(-10)(1^2) \]
This simplifies to:
\[ h = u - 5 \quad \text{(Equation 1)} \]

For \( t_2 = 7 \) seconds:
\[ h = u(7) + \frac{1}{2}(-10)(7^2) \]
This simplifies to:
\[ h = 7u - 245 \quad \text{(Equation 2)} \]

### Step 3: Equate the two expressions for \( h \)
Since both expressions equal \( h \), we can set them equal to each other:
\[ u - 5 = 7u - 245 \]

### Step 4: Solve for \( u \)
Rearranging the equation gives:
\[ -5 + 245 = 7u - u \]
\[ 240 = 6u \]
\[ u = \frac{240}{6} = 40 \, \text{m/s} \]

### Conclusion
The initial velocity \( u \) of the body is \( 40 \, \text{m/s} \).

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