A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground ?

Answer

To solve the problem of how much time elapses between the instant of release and the instant of impact of a stone thrown downward from a building, we can use the equations of motion. Here’s a step-by-step solution:

### Step 1: Identify the known values
- Initial velocity (u) = 20 m/s (downward, so we take it as negative: u = -20 m/s)
- Height of the building (s) = 60 m (downward, so we take it as negative: s = -60 m)
- Acceleration due to gravity (a) = 10 m/s² (downward, so we take it as negative: a = -10 m/s²)

### Step 2: Write the equation of motion
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ -60 = -20t + \frac{1}{2}(-10)t^2 \]

### Step 4: Simplify the equation
This simplifies to:
\[ -60 = -20t - 5t^2 \]

Rearranging gives:
\[ 5t^2 + 20t - 60 = 0 \]

### Step 5: Divide the entire equation by 5 to simplify
\[ t^2 + 4t - 12 = 0 \]

### Step 6: Factor the quadratic equation
To factor the quadratic equation:
\[ t^2 + 6t - 2t - 12 = 0 \]
This can be factored as:
\[ (t + 6)(t - 2) = 0 \]

### Step 7: Solve for t
Setting each factor to zero gives:
1. \( t + 6 = 0 \) → \( t = -6 \) (not a valid solution)
2. \( t - 2 = 0 \) → \( t = 2 \)

### Conclusion
The time that elapses between the instant of release and the instant of impact with the ground is **2 seconds**.

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