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)` [`g=10 ms^(-2)`]

Options

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

Answer

To solve the problem step-by-step, we can follow these logical steps:

### Step 1: Understand the Motion
A body is projected vertically upwards with an initial velocity \( U \). It crosses a certain height \( h \) twice at \( t = 1 \) second and \( t = 7 \) seconds.

### Step 2: Determine Total Time of Flight
The total time of flight can be deduced from the times at which the body crosses the height \( h \). Since it crosses the height at \( t = 1 \) second on the way up and at \( t = 7 \) seconds on the way down, the total time of flight \( T \) is:
\[ T = 1 + (T - 7) = 8 \text{ seconds} \]

### Step 3: Use the Time of Flight Formula
The formula for the total time of flight for a projectile launched vertically upwards is given by:
\[ T = \frac{2U}{g} \]
where \( g \) is the acceleration due to gravity. Given \( g = 10 \, \text{ms}^{-2} \), we can substitute this into the equation:
\[ 8 = \frac{2U}{10} \]

### Step 4: Solve for Initial Velocity \( U \)
Rearranging the equation to solve for \( U \):
\[ 8 = \frac{2U}{10} \implies 2U = 80 \implies U = 40 \, \text{ms}^{-1} \]

### Conclusion
The initial velocity \( U \) of the body is \( 40 \, \text{ms}^{-1} \).

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