Question

a brick is dropped from rest from a height of 4.9 m. how long does it take the brick to reach the ground?
2.0 s
1.4 s
1.0 s
0.6 s
1.2 s

Explanation:

Step1: Identify the kinematic - equation

The kinematic equation for vertical motion under gravity when the initial velocity $u = 0$ is $h=ut+\frac{1}{2}gt^{2}$. Since $u = 0$, the equation simplifies to $h=\frac{1}{2}gt^{2}$, where $h$ is the height, $g$ is the acceleration due to gravity ($g = 9.8\ m/s^{2}$), and $t$ is the time.

Step2: Rearrange the equation for time

From $h=\frac{1}{2}gt^{2}$, we can solve for $t$. First, multiply both sides by 2 to get $2h = gt^{2}$. Then, $t^{2}=\frac{2h}{g}$, and $t=\sqrt{\frac{2h}{g}}$.

Step3: Substitute the values

Given $h = 4.9\ m$ and $g = 9.8\ m/s^{2}$, we substitute these values into the formula: $t=\sqrt{\frac{2\times4.9}{9.8}}$.
$t=\sqrt{\frac{9.8}{9.8}}=\sqrt{1}=1.0\ s$.

Answer:

C. 1.0 s