A platform is moving upwards with an accelerations of `5ms^(-2)`. At the moment when its velocity is ` u = 3ms^(-1)`, a ball is thrown from it with a speed of `30 ms^(-1)` w.r.t. platform at an angle of `theta = 30^@` with horizontal. The time taken by the ball to return to the platform is

Options

A. 2 s
B. 3 s
C. 1 s
D. `2.5 s`

Explanation

To solve the problem, we need to determine the time taken by the ball to return to the moving platform after being thrown. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
The platform is moving upwards with an acceleration of \( a = 5 \, \text{m/s}^2 \) and has an initial velocity of \( u = 3 \, \text{m/s} \). A ball is thrown from the platform with a speed of \( 30 \, \text{m/s} \) at an angle of \( \theta = 30^\circ \) with respect to the horizontal.

### Step 2: Resolve the Initial Velocity of the Ball
The initial velocity of the ball can be resolved into horizontal and vertical components. The vertical component of the velocity \( V_{y} \) is given by:
\[ V_{y} = 30 \sin(30^\circ) = 30 \times \frac{1}{2} = 15 \, \text{m/s} \]

### Step 3: Determine the Effective Acceleration of the Ball
The ball is under the influence of gravity \( g \) acting downwards, which we can take as \( g = 10 \, \text{m/s}^2 \). The platform is accelerating upwards with \( a = 5 \, \text{m/s}^2 \). Thus, the effective acceleration \( A \) acting on the ball with respect to the platform is:
\[ A = -g + a = -10 + 5 = -5 \, \text{m/s}^2 \]
This means the ball experiences a net downward acceleration of \( 5 \, \text{m/s}^2 \) relative to the platform.

### Step 4: Apply the Kinematic Equation
We will use the kinematic equation for vertical motion to find the time \( t \) when the ball returns to the platform. The displacement \( S_y \) of the ball when it returns to the platform is zero:
\[ S_y = V_{y} t + \frac{1}{2} A t^2 \]
Substituting the known values:
\[ 0 = 15t + \frac{1}{2} (-5) t^2 \]
This simplifies to:
\[ 0 = 15t - \frac{5}{2} t^2 \]
Factoring out \( t \):
\[ t(15 - \frac{5}{2}t) = 0 \]
This gives us two solutions:
\[ t = 0 \quad \text{or} \quad 15 - \frac{5}{2}t = 0 \]

### Step 5: Solve for \( t \)
From the second equation:
\[ 15 = \frac{5}{2}t \]
Multiplying both sides by \( 2 \):
\[ 30 = 5t \]
Dividing by \( 5 \):
\[ t = 6 \, \text{s} \]

### Step 6: Check the Validity of the Solution
Since \( t = 0 \) corresponds to the moment the ball is thrown, the only valid solution is \( t = 6 \, \text{s} \).

### Final Answer
The time taken by the ball to return to the platform is \( t = 6 \, \text{s} \).

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