practice a\n1. how long was the rock in the air?\n2. what was the rocks velocity when it hit the…

practice a\n1. how long was the rock in the air?\n2. what was the rocks velocity when it hit the ground\ngiven: hinahanap:\na = -9.8 m/s² t =?\nvi = 0 m/s vf =?\nd = 120 m

practice a\n1. how long was the rock in the air?\n2. what was the rocks velocity when it hit the ground\ngiven: hinahanap:\na = -9.8 m/s² t =?\nvi = 0 m/s vf =?\nd = 120 m

Answer

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 $- 48.51\ m/s$.