A plane is flying horizontally at `98ms^(-1)` and releases and object which reaches the ground in 10s. The angle made by it while hitting th ground is:-

Options

A. `55^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem step by step, we will analyze the motion of the object released from the plane and calculate the angle it makes with the ground upon impact.

### Step 1: Identify the horizontal and vertical components of motion
The plane is flying horizontally at a speed of \( v_x = 98 \, \text{m/s} \). When the object is released, it has no initial vertical velocity (\( u_y = 0 \)). The only force acting on it after release is gravity, which accelerates it downwards at \( g = 9.8 \, \text{m/s}^2 \).

### Step 2: Calculate the vertical velocity just before impact
Using the equation of motion for the vertical component:
\[ v_y = u_y + a_y t \]
Here, \( u_y = 0 \), \( a_y = g = 9.8 \, \text{m/s}^2 \), and \( t = 10 \, \text{s} \):
\[ v_y = 0 + 9.8 \times 10 = 98 \, \text{m/s} \]

### Step 3: Determine the angle of impact
The angle \( \theta \) made by the object with the ground can be found using the tangent function:
\[ \tan \theta = \frac{v_y}{v_x} \]
Substituting the values we found:
\[ \tan \theta = \frac{98 \, \text{m/s}}{98 \, \text{m/s}} = 1 \]

### Step 4: Calculate the angle \( \theta \)
The angle \( \theta \) can be found using the inverse tangent function:
\[ \theta = \tan^{-1}(1) \]
This gives:
\[ \theta = 45^\circ \]

### Final Answer
The angle made by the object while hitting the ground is \( 45^\circ \).

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