A ball is thrown upwards from the ground with an initial speed of u. The ball is at height of `80m` at two times, the time interval being 6 s. Find u. Take `g=10 m//s^2.`

Explanation

To solve the problem step by step, we will use the kinematic equations of motion. The key points to note are that the ball reaches a height of 80 m at two different times, and the time interval between these two instances is 6 seconds. We will also consider the acceleration due to gravity (g) as -10 m/s² (negative because it acts downward).

### Step-by-Step Solution:

1. **Identify the known values**:
- Height (s) = 80 m
- Time interval (Δt) = 6 s
- Acceleration due to gravity (g) = -10 m/s²

2. **Use the kinematic equation**:
The kinematic equation we will use is:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the displacement (80 m),
- \( u \) is the initial velocity (which we need to find),
- \( a \) is the acceleration (-10 m/s²),
- \( t \) is the time.

Plugging in the values, we get:
\[ 80 = ut - 5t^2 \]
Rearranging gives us:
\[ 5t^2 - ut + 80 = 0 \]

3. **Recognize the quadratic equation**:
The equation \( 5t^2 - ut + 80 = 0 \) is a quadratic equation in \( t \). The roots of this equation will give us the two times (t1 and t2) at which the ball is at a height of 80 m.

4. **Use the quadratic formula**:
The roots of the quadratic equation \( at^2 + bt + c = 0 \) can be found using:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -u \), and \( c = 80 \). Thus, we have:
\[ t = \frac{u \pm \sqrt{u^2 - 1600}}{10} \]

5. **Set up the time difference**:
The difference between the two times (t2 - t1) is given as 6 seconds:
\[ t_2 - t_1 = \frac{u + \sqrt{u^2 - 1600}}{10} - \frac{u - \sqrt{u^2 - 1600}}{10} = 6 \]
Simplifying this gives:
\[ \frac{2\sqrt{u^2 - 1600}}{10} = 6 \]
Multiplying both sides by 10:
\[ 2\sqrt{u^2 - 1600} = 60 \]
Dividing by 2:
\[ \sqrt{u^2 - 1600} = 30 \]

6. **Square both sides**:
Squaring both sides gives:
\[ u^2 - 1600 = 900 \]
Rearranging gives:
\[ u^2 = 2500 \]

7. **Solve for u**:
Taking the square root of both sides:
\[ u = \sqrt{2500} = 50 \text{ m/s} \]

### Final Answer:
The initial speed \( u \) at which the ball was thrown upwards is \( 50 \text{ m/s} \).

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