A particle `P` is projected upwards with `80m//s`. One second later another particle `Q` is projected with initial velocity `70m//s`. Before either of the particle srikes the ground `(g=10m//s^(2))`

Options

A. both particle are at rest with respect to each other
B. after `2s` distance between the particles is `75m`
C. when particle `P` is at highest point, particle `Q` is moving downwards
D. when particle `P` is at highest point, particle `Q` is moving upwards

Answer

To solve the problem, we need to analyze the motion of both particles, P and Q, step by step.

### Step 1: Determine the motion of Particle P
Particle P is projected upwards with an initial velocity of \( u_P = 80 \, \text{m/s} \). The time of flight until it reaches the highest point can be calculated using the formula:
\[ v = u - gt \]
At the highest point, the final velocity \( v = 0 \):
\[ 0 = 80 - 10t_P \implies t_P = \frac{80}{10} = 8 \, \text{s} \]
So, Particle P will take 8 seconds to reach the highest point.

### Step 2: Calculate the maximum height reached by Particle P
Using the formula for displacement:
\[ s_P = ut - \frac{1}{2}gt^2 \]
Substituting \( u = 80 \, \text{m/s} \), \( g = 10 \, \text{m/s}^2 \), and \( t = 8 \, \text{s} \):
\[ s_P = 80 \times 8 - \frac{1}{2} \times 10 \times (8^2) = 640 - 320 = 320 \, \text{m} \]
So, Particle P reaches a maximum height of 320 meters.

### Step 3: Determine the motion of Particle Q
Particle Q is projected 1 second after Particle P with an initial velocity of \( u_Q = 70 \, \text{m/s} \). Therefore, by the time Particle Q is projected, Particle P has already been in motion for 1 second.

### Step 4: Calculate the height of Particle P after 1 second
Using the same displacement formula for Particle P at \( t = 1 \, \text{s} \):
\[ s_P(1) = 80 \times 1 - \frac{1}{2} \times 10 \times (1^2) = 80 - 5 = 75 \, \text{m} \]
After 1 second, Particle P is at a height of 75 meters.

### Step 5: Calculate the time of flight for Particle Q
Now, Particle Q will be in motion for \( t_Q = 8 - 1 = 7 \, \text{s} \) (it will take 7 seconds to reach the highest point).

### Step 6: Calculate the maximum height reached by Particle Q
Using the same formula:
\[ s_Q = u_Q t_Q - \frac{1}{2} g t_Q^2 \]
Substituting \( u_Q = 70 \, \text{m/s} \) and \( t_Q = 7 \, \text{s} \):
\[ s_Q = 70 \times 7 - \frac{1}{2} \times 10 \times (7^2) = 490 - 245 = 245 \, \text{m} \]
So, Particle Q reaches a maximum height of 245 meters.

### Step 7: Determine the total height of both particles
After 2 seconds, Particle P is at 75 meters, and Particle Q is at 0 meters (as it has just been projected).

### Step 8: Determine the time taken for both particles to hit the ground
- Particle P will take a total of \( 8 \, \text{s} \) to reach the ground after reaching its maximum height.
- Particle Q will take \( 7 \, \text{s} \) to reach its maximum height and then will take the same time to fall back down, totaling \( 7 + 7 = 14 \, \text{s} \).

### Conclusion
Both particles will hit the ground at different times, with Particle P hitting the ground first after 8 seconds, and Particle Q hitting the ground later.