Question

practice a

1. how long was the rock in the air?
2. what was the rocks velocity when it hit the ground

given: hinahanap:
a = -9.8 m/s² t =?
vi = 0 m/s vf =?
d = 120 m

Explanation:

Step1: Use the displacement - time formula

We use the formula $d = v_i t+\frac{1}{2}at^{2}$. Since $v_i = 0\ m/s$, the formula simplifies to $d=\frac{1}{2}at^{2}$.
$120=\frac{1}{2}\times9.8\times t^{2}$ (taking magnitude of $a = 9.8\ m/s^{2}$ as we are dealing with magnitudes for now).

Step2: Solve for time $t$

First, rewrite the equation as $t^{2}=\frac{2\times120}{9.8}$.
$t^{2}=\frac{240}{9.8}\approx24.49$.
Then $t=\sqrt{24.49}\approx 4.95\ s$.

Step3: Use the velocity - time formula

To find the final velocity $v_f$, we use the formula $v_f=v_i + at$.
Since $v_i = 0\ m/s$ and $a=- 9.8\ m/s^{2}$ and $t = 4.95\ s$.
$v_f=0+( - 9.8)\times4.95=-48.51\ m/s$. The negative sign indicates the direction is downwards.

Answer:

1. The rock was in the air for approximately $4.95\ s$.
2. The rock's velocity when it hit the ground was approximately $- 48.51\ m/s$.