A body P is thrown vertically up with velocity `30 ms^(-1)` and another body Q is thrown up along the same vertically line with the same velocity but 1 second later from the ground. When they meet `(g = 10 ms-2)`

Options

A. P travels for 2.5 s
B. Q travels for 3.5 s
C. P travels for 3.5 s
D. Q travels for 1 s

Answer

To solve the problem, we need to analyze the motion of both bodies P and Q. Let's break it down step by step.

### Step 1: Define the variables
Let:
- \( u = 30 \, \text{m/s} \) (initial velocity of both bodies)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t \) = time taken by body P until they meet
- \( t' = t - 1 \) = time taken by body Q until they meet (since Q is thrown 1 second later)

### Step 2: Write the equations of motion
For body P, which is thrown upwards:
\[ S_P = ut - \frac{1}{2}gt^2 \]
For body Q, which is thrown upwards after 1 second:
\[ S_Q = ut' - \frac{1}{2}gt'^2 \]

### Step 3: Substitute \( t' \) in the equation for Q
Since \( t' = t - 1 \):
\[ S_Q = u(t - 1) - \frac{1}{2}g(t - 1)^2 \]

### Step 4: Set the displacements equal
Since both bodies meet at the same height, we can equate \( S_P \) and \( S_Q \):
\[ ut - \frac{1}{2}gt^2 = u(t - 1) - \frac{1}{2}g(t - 1)^2 \]

### Step 5: Expand the equation for Q
Expanding \( S_Q \):
\[ S_Q = ut - u - \frac{1}{2}g(t^2 - 2t + 1) \]
\[ = ut - u - \frac{1}{2}gt^2 + gt - \frac{1}{2}g \]

### Step 6: Combine the equations
Now, equate the two expressions:
\[ ut - \frac{1}{2}gt^2 = ut - u - \frac{1}{2}gt^2 + gt - \frac{1}{2}g \]

### Step 7: Simplify the equation
Cancel \( ut \) and \( -\frac{1}{2}gt^2 \) from both sides:
\[ 0 = -u + gt - \frac{1}{2}g \]
Rearranging gives:
\[ gt = u + \frac{1}{2}g \]

### Step 8: Substitute the values
Substituting \( u = 30 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \):
\[ gt = 30 + \frac{1}{2} \times 10 \]
\[ gt = 30 + 5 = 35 \]

### Step 9: Solve for \( t \)
Now, divide by \( g \):
\[ t = \frac{35}{10} = 3.5 \, \text{s} \]

### Conclusion
The total time that body P travels before meeting body Q is \( 3.5 \, \text{s} \).