A ball is thrown upwards from the top of a tower `40 m` high with a velocity of `10 m//s.` Find the time when it strikes the ground. Take `g=10 m//s^2`.

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Identify the known values
- Height of the tower (s) = -40 m (negative because it's downward)
- Initial velocity (u) = +10 m/s (upward)
- Acceleration due to gravity (g) = -10 m/s² (downward)

### Step 2: Use the kinematic equation
We will use the kinematic equation:
\[ s = ut + \frac{1}{2} a t^2 \]
Where:
- \( s \) = displacement
- \( u \) = initial velocity
- \( a \) = acceleration
- \( t \) = time

Substituting the known values into the equation:
\[ -40 = 10t + \frac{1}{2}(-10)t^2 \]

### Step 3: Simplify the equation
This simplifies to:
\[ -40 = 10t - 5t^2 \]
Rearranging gives us:
\[ 5t^2 - 10t - 40 = 0 \]

### Step 4: Divide through by 5
To simplify further, divide the entire equation by 5:
\[ t^2 - 2t - 8 = 0 \]

### Step 5: Factor the quadratic equation
Now we will factor the quadratic equation:
\[ (t - 4)(t + 2) = 0 \]

### Step 6: Solve for t
Setting each factor to zero gives us:
1. \( t - 4 = 0 \) → \( t = 4 \) seconds
2. \( t + 2 = 0 \) → \( t = -2 \) seconds (not physically meaningful)

### Step 7: Conclusion
The only valid solution is:
\[ t = 4 \text{ seconds} \]

Thus, the time when the ball strikes the ground is **4 seconds**.

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