A ball is dropped from a window of 19.62 meters high. How long will it take to reach the ground?
2 sec
Physics Problem: Calculating Fall Time for a Dropped Ball
This problem involves calculating the time it takes for an object to fall under the influence of gravity, a concept known as free fall. We are given the height from which the ball is dropped and need to find the duration of its fall.
Understanding Free Fall Physics
When an object is dropped, it starts with zero initial velocity and accelerates downwards due to gravity. The standard acceleration due to gravity on Earth is approximately $g = 9.81 \, \text{m/s}^2$. We can use the equations of motion (kinematics) to solve this problem.
Given Information
Let's list the known values:
- Height ($h$): $19.62$ meters
- Initial Velocity ($u$): $0 \, \text{m/s}$ (since the ball is dropped)
- Acceleration ($a$): $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity)
- Time ($t$): This is what we need to find.
Relevant Kinematic Equation
The equation that relates distance, initial velocity, acceleration, and time is:
$$ s = ut + \frac{1}{2}at^2 $$In this context, the distance $s$ is the height $h$, and the acceleration $a$ is gravity $g$. So the equation becomes:
$$ h = ut + \frac{1}{2}gt^2 $$Step-by-Step Calculation
1. Substitute the known values into the equation:
$$ 19.62 = (0 \times t) + \frac{1}{2}(9.81)t^2 $$2. Simplify the equation:
$$ 19.62 = 0 + \frac{1}{2}(9.81)t^2 $$ $$ 19.62 = 4.905 t^2 $$3. Solve for $t^2$ by dividing both sides by $4.905$:
$$ t^2 = \frac{19.62}{4.905} $$ $$ t^2 = 4 $$4. Find the time $t$ by taking the square root of both sides:
$$ t = \sqrt{4} $$ $$ t = 2 \, \text{seconds} $$Summary of Results
Here's a summary of the parameters used and the calculated result:
| Parameter | Value |
|---|---|
| Height ($h$) | $19.62$ m |
| Initial Velocity ($u$) | $0$ m/s |
| Acceleration ($g$) | $9.81$ m/s$^2$ |
| Time ($t$) | $2$ sec |
Conclusion
The calculation shows that it will take the ball 2 seconds to reach the ground when dropped from a height of 19.62 meters, assuming negligible air resistance and using $g = 9.81 \, \text{m/s}^2$. This matches one of the provided options.
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