Free fall kinematics: 3 worked examples

To determine the distance a rock falls from rest under the influence of gravity, we use the kinematic equation for displacement under constant acceleration:
$d = v_{i}t + \frac{1}{2}gt^{2}$
Where:
- $d$ is the distance fallen.
- $v_{i}$ is the initial velocity, which is $0 \text{ m/s}$ because the rock is dropped from rest.
- $g$ is the acceleration due to gravity, given as $9.8 \text{ m/s}^{2}$.
- $t$ is the time elapsed, which is $2.0 \text{ s}$.

Plugging in the values:
$d = (0 \text{ m/s})(2.0 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s}^{2})(2.0 \text{ s})^{2}$
$d = 0 + \frac{1}{2}(9.8 \text{ m/s}^{2})(4.0 \text{ s}^{2})$
$d = (4.9 \text{ m/s}^{2})(4.0 \text{ s}^{2})$
$d = 19.6 \text{ m}$

To find the maximum height reached by the ball, we can use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
$v_{f}^{2} = v_{i}^{2} + 2a\Delta y$

In this scenario:

1. The initial velocity $v_{i}$ is $15 \text{ m/s}$.
2. At the maximum height, the final velocity $v_{f}$ is $0 \text{ m/s}$.
3. The acceleration $a$ is due to gravity and acts downward, so $a = -g = -9.8 \text{ m/s}^{2}$.
4. The displacement $\Delta y$ represents the maximum height $h$.

Plugging these values into the equation:
$0^{2} = (15)^{2} + 2(-9.8)h$
$0 = 225 - 19.6h$

Now, solve for $h$:
$19.6h = 225$
$h = \frac{225}{19.6}$
$h \approx 11.47959...$

Rounding to two decimal places, the maximum height is approximately $11.48 \text{ m}$.

To find the time $t$ it takes for an object dropped from a height $h$ to reach the ground, we use the kinematic equation for free fall:
$h = \frac{1}{2}gt^{2}$
In this problem, the height $h = 45$ m and the acceleration due to gravity $g = 10$ m/s$^2$.
Substituting these values into the equation:
$45 = \frac{1}{2}(10)t^{2}$
$45 = 5t^{2}$
To solve for $t^2$, divide both sides by 5:
$t^{2} = \frac{45}{5}$
$t^{2} = 9$
Taking the square root of both sides, we find:
$t = \sqrt{9}$
$t = 3$ s