A body is projected horizontally from the top of a tower with a velocity of `10 m//s`.If it hits the ground at an angle `45^(@)`, then vertical component of velocity when it hits ground in `m//s` is

Options

A. `10 sqrt(2)`
B. `5sqrt(2)`
C. 5
D. 10

Explanation

To solve the problem step by step, we need to determine the vertical component of the velocity (Vy) when the body hits the ground at an angle of 45 degrees.

### Step 1: Understand the Given Information
- The body is projected horizontally with an initial horizontal velocity (Vx) of 10 m/s.
- It hits the ground at an angle of 45 degrees.

### Step 2: Analyze the Motion
Since the body is projected horizontally, the initial vertical velocity (Vy_initial) is 0 m/s. The only force acting on the body in the vertical direction is gravity, which will cause the body to accelerate downward.

### Step 3: Use the Angle of Impact
When the body hits the ground at an angle of 45 degrees, we can use the relationship between the vertical and horizontal components of the velocity. The tangent of the angle (θ) is given by:
\[ \tan(\theta) = \frac{Vy}{Vx} \]
For θ = 45 degrees, we know that:
\[ \tan(45^\circ) = 1 \]
Thus, we have:
\[ 1 = \frac{Vy}{Vx} \]

### Step 4: Substitute the Known Values
We know that Vx = 10 m/s (the horizontal component of the velocity remains constant since there is no horizontal acceleration). Therefore, we can substitute this into the equation:
\[ 1 = \frac{Vy}{10} \]

### Step 5: Solve for Vy
From the equation above, we can solve for Vy:
\[ Vy = 10 \times 1 = 10 \text{ m/s} \]

### Conclusion
The vertical component of the velocity when the body hits the ground is **10 m/s**.

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