When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 metres.
(a) What was the initial speed of the ball ?
(b) How much time is taken by the ball to reach the highest point ? `(g=10 m s^(-2))`

Answer

To solve the problem step by step, we will use the equations of motion under uniform acceleration due to gravity.

### Given:
- Maximum height (s) = 5 m
- Acceleration due to gravity (g) = 10 m/s² (acting downwards)
- Final velocity (v) at the maximum height = 0 m/s (the ball stops momentarily at the highest point)

### (a) Finding the initial speed (u) of the ball:

1. **Use the equation of motion:**
\[ v^2 = u^2 + 2as \]
Here, \( a \) is the acceleration. Since gravity acts downwards, we take \( a = -g = -10 \, \text{m/s}^2 \).

2. **Substituting the known values:**
\[ 0^2 = u^2 + 2(-10)(5) \]
\[ 0 = u^2 - 100 \]

3. **Rearranging the equation:**
\[ u^2 = 100 \]

4. **Taking the square root:**
\[ u = \sqrt{100} = 10 \, \text{m/s} \]

### (b) Finding the time taken (t) to reach the highest point:

1. **Use the equation of motion:**
\[ v = u + at \]
Again, \( a = -g = -10 \, \text{m/s}^2 \).

2. **Substituting the known values:**
\[ 0 = 10 + (-10)t \]
\[ 0 = 10 - 10t \]

3. **Rearranging the equation:**
\[ 10t = 10 \]

4. **Solving for t:**
\[ t = \frac{10}{10} = 1 \, \text{s} \]

### Final Answers:
(a) The initial speed of the ball is **10 m/s**.
(b) The time taken by the ball to reach the highest point is **1 second**.