Question

derive simplified equations that you could use to calculate an objects displacement and final velocity when it starts at rest a height (h) above the ground and then goes into free - fall at constant acceleration. begin with the generic equations for displacement ($x_f=x_i + v_it+left(\frac{1}{2}
ight)vec{a}t^{2}$) and velocity ($v_f = v_i+at$). assume the positive axis is upward and that this scenario occurs near earths surface where $g = 9.8 m/s^{2}$ downward. use the equations you just derived to find the velocity and position of a ball that was dropped from a height of 10.0 m after it has fallen for 1.0 s.

Explanation:

Step1: Identify initial conditions

The object starts at rest, so $v_i = 0$. Let the initial height be $h$, so $x_i=h$. The acceleration $a=-g=- 9.8\ m/s^{2}$ (negative because it's downward and positive - axis is upward).

Step2: Simplify displacement equation

The general displacement equation is $x_f=x_i + v_i t+\frac{1}{2}at^{2}$. Substituting $v_i = 0$ and $a=-g$, we get $x_f=h-\frac{1}{2}gt^{2}$.

Step3: Simplify velocity equation

The general velocity equation is $v_f = v_i+at$. Substituting $v_i = 0$ and $a = - g$, we get $v_f=-gt$.

Step4: Calculate displacement and velocity for given values

Given $h = 10.0\ m$, $t = 1.0\ s$ and $g = 9.8\ m/s^{2}$.
For displacement: $x_f=10-\frac{1}{2}\times9.8\times(1)^{2}=10 - 4.9=5.1\ m$.
For velocity: $v_f=-9.8\times1=-9.8\ m/s$.

Answer:

Displacement equation: $x_f=h-\frac{1}{2}gt^{2}$
Velocity equation: $v_f=-gt$
Displacement at $h = 10.0\ m$, $t = 1.0\ s$: $5.1\ m$
Velocity at $h = 10.0\ m$, $t = 1.0\ s$: $- 9.8\ m/s$