A ball is projected horizontallu from a top of tower with a velocity of `10 ms^(-1)`. Find the velocity of the ball after `0.5 s. Take ` g= 10 ms^(-2)`.

Explanation

To find the velocity of the ball after 0.5 seconds when it is projected horizontally from the top of a tower, we can break the problem down into horizontal and vertical components.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The ball is projected horizontally with an initial velocity \( u_x = 10 \, \text{m/s} \).
- The initial vertical velocity \( u_y = 0 \, \text{m/s} \) (since it is projected horizontally).
- The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards.
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