Question

free fall & graphs of motion
determine the velocity of a stone, immediately before it hits the ground. if it is dropped from a height of 20 m.
use g = -10 m/s².
$v_f=v_0 + gt$
$delta y=v_0t+\frac{at^{2}}{2}$
$v_{f}^{2}=v_{0}^{2}+2gdelta y$
-20 m/s
20 m/s
-10 m/s
10 m/s

Explanation:

Step1: Identify the kinematic - equation

We use the equation $v_f^2=v_0^2 + 2g\Delta y$. Since the stone is dropped, $v_0 = 0\ m/s$, $g=- 10\ m/s^2$ and $\Delta y=-20\ m$.

Step2: Substitute the values

Substitute $v_0 = 0$, $g=-10\ m/s^2$ and $\Delta y=-20\ m$ into the equation $v_f^2=v_0^2 + 2g\Delta y$. We get $v_f^2=0+2\times(- 10)\times(-20)$.

Step3: Calculate $v_f^2$

$v_f^2=400\ (m/s)^2$.

Step4: Solve for $v_f$

Taking the square - root of both sides, $v_f=\pm20\ m/s$. Since the stone is falling downwards, the velocity is negative. So $v_f=-20\ m/s$.

Answer:

$-20\ m/s$