A cart is moving alongs +ve x-direction with a speed of 4m/s Aperson on the cart throw a stone with a velcity of 6 m/s w.r.t hiumself In the frame of refrence of the cart the stone is thrown in the y-z palne making an angle of `30^(@)` with the verticle z- axies . Then w.r.t an observer on the ground

Options

A. intial speed of the stone is 10 m/s
B. the intial speed of the stone is `2sqrt(13)`m/s
C. the velocity at the highest points of its motion is zero
D. the velocity at the highest points of its motion is 5 m/s

Explanation

To solve the problem, we need to analyze the motion of the stone thrown from the cart in the context of different reference frames. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the motion of the cart and the stone
- The cart is moving in the positive x-direction with a velocity of \( V_c = 4 \, \text{m/s} \).
- The stone is thrown by a person on the cart with a velocity of \( V_s = 6 \, \text{m/s} \) relative to the cart.

### Step 2: Analyze the angle of projection
- The stone is thrown in the y-z plane, making an angle of \( 30^\circ \) with the vertical z-axis.
- This means we can resolve the stone's velocity into its y and z components:
- The vertical component (z-direction) is given by:
\[ V_z = V_s \cos(30^\circ) = 6 \cos(30^\circ) = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} \, \text{m/s} \]
- The horizontal component (y-direction) is given by:
\[ V_y = V_s \sin(30^\circ) = 6 \sin(30^\circ) = 6 \times \frac{1}{2} = 3 \, \text{m/s} \]

### Step 3: Combine the velocities in the ground frame
- The stone's velocity in the ground frame will have three components: x, y, and z.
- The x-component of the stone's velocity is simply the velocity of the cart:
\[ V_x = V_c = 4 \, \text{m/s} \]
- Therefore, the total velocity components of the stone with respect to an observer on the ground are:
- \( V_x = 4 \, \text{m/s} \)
- \( V_y = 3 \, \text{m/s} \)
- \( V_z = 3\sqrt{3} \, \text{m/s} \)

### Step 4: Calculate the resultant velocity of the stone
- The magnitude of the resultant velocity \( V \) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2 + V_z^2} \]
Substituting the values:
\[ V = \sqrt{(4)^2 + (3)^2 + (3\sqrt{3})^2} \]
\[ V = \sqrt{16 + 9 + 27} = \sqrt{52} = 2\sqrt{13} \, \text{m/s} \]

### Step 5: Determine the velocity at the highest point
- At the highest point of its trajectory, the vertical component of the velocity (z-direction) will be zero due to gravity acting on the stone.
- The horizontal components will remain unchanged:
- \( V_x = 4 \, \text{m/s} \)
- \( V_y = 3 \, \text{m/s} \)

### Step 6: Calculate the speed at the highest point
- The speed at the highest point can be calculated as:
\[ V_{top} = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V_{top} = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{m/s} \]

### Final Result
- The resultant speed of the stone with respect to an observer on the ground is \( 2\sqrt{13} \, \text{m/s} \).
- The speed of the stone at the highest point is \( 5 \, \text{m/s} \).