A ball is gently dropped from a height of `20m`. If its velocity increases uniformly at the rate of `10 m//s^(2)`, with what velocity will it strike the ground ? After what time will it strike the ground ?

Explanation

To solve the problem step by step, we will use the equations of motion.

**Step 1: Identify the known values.**
- Initial velocity (u) = 0 m/s (since the ball is dropped)
- Acceleration (a) = 10 m/s² (given)
- Height (h) = 20 m

**Step 2: Use the third equation of motion to find the final velocity (v).**
The third equation of motion is:
\[ v^2 = u^2 + 2as \]
Here, we can replace \( s \) with \( h \) (the height from which the ball is dropped).

Substituting the known values:
\[ v^2 = 0^2 + 2 \times 10 \times 20 \]
\[ v^2 = 0 + 400 \]
\[ v^2 = 400 \]

Now, take the square root to find \( v \):
\[ v = \sqrt{400} \]
\[ v = 20 \, \text{m/s} \]

**Step 3: Use the first equation of motion to find the time (t).**
The first equation of motion is:
\[ v = u + at \]

Substituting the known values:
\[ 20 = 0 + 10t \]

Now, solve for \( t \):
\[ 20 = 10t \]
\[ t = \frac{20}{10} \]
\[ t = 2 \, \text{seconds} \]

**Final Answers:**
- The velocity with which the ball strikes the ground is **20 m/s**.
- The time taken to strike the ground is **2 seconds**.

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