Two particle A and B projected along different directions from the same point P on the ground with the same velocity of 70 m/s in the same vertical plane. They hit the ground at the same point Q such that PQ = 480,. Then `[g=9.8 m//s^(2)]`

Options

A. Ratio of their times of flight is `4 : 5`
B. Ratio of their maximum heights is `9 : 16`
C. Ratio of their minimum speeds during flights is `4 : 3`
D. The bisector of the angle between their directions of projection makes `45^(@)` with horizontal

Answer

To solve the problem step by step, we will analyze the projectile motion of two particles A and B, which are projected from the same point P with the same initial velocity and hit the ground at the same point Q.

### Step 1: Understanding the Problem
We have two particles A and B projected at the same initial velocity \( u = 70 \, \text{m/s} \) at angles \( \theta_A \) and \( \theta_B \) respectively. They hit the ground at the same horizontal distance \( PQ = 480 \, \text{m} \). The acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \).

### Step 2: Range of Projectile Motion
The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Since both particles hit the same point Q, their ranges are equal:
\[ R_A = R_B = 480 \, \text{m} \]
Thus, we can write:
\[ 480 = \frac{70^2 \sin(2\theta_A)}{9.8} \]
\[ 480 = \frac{70^2 \sin(2\theta_B)}{9.8} \]

### Step 3: Solving for Angles
From the range formula for particle A:
\[ 480 = \frac{4900 \sin(2\theta_A)}{9.8} \]
Rearranging gives:
\[ \sin(2\theta_A) = \frac{480 \times 9.8}{4900} \]
Calculating this:
\[ \sin(2\theta_A) = \frac{4704}{4900} \approx 0.96 \]

Now, we can find \( 2\theta_A \):
\[ 2\theta_A = \sin^{-1}(0.96) \approx 73.74^\circ \]
Thus,
\[ \theta_A \approx 36.87^\circ \]

Since \( \theta_A + \theta_B = 90^\circ \):
\[ \theta_B = 90^\circ - \theta_A \approx 53.13^\circ \]

### Step 4: Time of Flight
The time of flight \( T \) for projectile motion is given by:
\[ T = \frac{2u \sin(\theta)}{g} \]
Thus, for particles A and B:
\[ T_A = \frac{2 \times 70 \sin(36.87^\circ)}{9.8} \]
\[ T_B = \frac{2 \times 70 \sin(53.13^\circ)}{9.8} \]

Calculating the sine values:
\[ \sin(36.87^\circ) \approx 0.6 \quad \text{and} \quad \sin(53.13^\circ) \approx 0.8 \]
Thus:
\[ T_A \approx \frac{2 \times 70 \times 0.6}{9.8} \approx 8.57 \, \text{s} \]
\[ T_B \approx \frac{2 \times 70 \times 0.8}{9.8} \approx 10.14 \, \text{s} \]

### Step 5: Maximum Height
The maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
Calculating for both particles:
\[ H_A = \frac{70^2 \sin^2(36.87^\circ)}{2 \times 9.8} \]
\[ H_B = \frac{70^2 \sin^2(53.13^\circ)}{2 \times 9.8} \]

Calculating:
\[ H_A \approx \frac{4900 \times 0.36}{19.6} \approx 90.0 \, \text{m} \]
\[ H_B \approx \frac{4900 \times 0.64}{19.6} \approx 160.0 \, \text{m} \]

### Step 6: Minimum Speed During Flight
The minimum speed occurs at the maximum height, which is given by the horizontal component of the velocity:
\[ V_{min} = u \cos(\theta) \]
Calculating for both particles:
\[ V_{minA} = 70 \cos(36.87^\circ) \approx 70 \times 0.8 = 56 \, \text{m/s} \]
\[ V_{minB} = 70 \cos(53.13^\circ) \approx 70 \times 0.6 = 42 \, \text{m/s} \]

### Step 7: Conclusion
Based on the calculations:
- The angles of projection are approximately \( 36.87^\circ \) and \( 53.13^\circ \).
- The ratio of time of flight \( T_A : T_B \) is approximately \( 3:4 \).
- The ratio of maximum heights \( H_A : H_B \) is \( 9:16 \).
- The ratio of minimum speeds \( V_{minA} : V_{minB} \) is \( 4:3 \).