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^(@)`

Explanation

To solve the problem step by step, we need to analyze the motion of the object released from the plane.

### Step 1: Identify the initial conditions
The plane is flying horizontally at a speed of \( V_x = 98 \, \text{m/s} \). The object is released from the plane and falls for \( t = 10 \, \text{s} \).

### Step 2: Calculate the vertical velocity component when it hits the ground
Since the object is dropped, its initial vertical velocity \( V_{y0} = 0 \). The vertical acceleration \( a_y \) is due to gravity, which is approximately \( 9.8 \, \text{m/s}^2 \). We can use the following kinematic equation to find the final vertical velocity \( V_y \):

\[ V_y = V_{y0} + a_y \cdot t \]

Substituting the known values:

\[ V_y = 0 + 9.8 \cdot 10 = 98 \, \text{m/s} \]

### Step 3: Determine the horizontal velocity component
The horizontal velocity \( V_x \) remains constant throughout the motion because there is no horizontal acceleration. Therefore:

\[ V_x = 98 \, \text{m/s} \]

### Step 4: Calculate the angle of impact
The angle \( \theta \) made by the object with the horizontal when it hits the ground can be found using the tangent function:

\[ \tan(\theta) = \frac{V_y}{V_x} \]

Substituting the values we calculated:

\[ \tan(\theta) = \frac{98}{98} = 1 \]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), we take the arctangent:

\[ \theta = \tan^{-1}(1) = 45^\circ \]

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

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