a ball is initially at rest and falls a distance of δy to the ground. the ball has an acceleration of \g\…

a ball is initially at rest and falls a distance of δy to the ground. the ball has an acceleration of \g\. which equation below could be used to find the time for the ball to fall to the ground below in terms of the given variables?\n$gt^{2}$\n$sqrt{\frac{g}{2}}$\n$sqrt{\frac{2delta y}{g}}$\n$\frac{1}{2}gt^{2}$\n$-\frac{g^{2}}{2t}$

a ball is initially at rest and falls a distance of δy to the ground. the ball has an acceleration of \g\. which equation below could be used to find the time for the ball to fall to the ground below in terms of the given variables?\n$gt^{2}$\n$sqrt{\frac{g}{2}}$\n$sqrt{\frac{2delta y}{g}}$\n$\frac{1}{2}gt^{2}$\n$-\frac{g^{2}}{2t}$

Answer

Explanation:

Step1: Recall kinematic - equation

The kinematic equation for an object in free - fall starting from rest ($v_0 = 0$) is $\Delta y=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$ and $a = g$, the equation simplifies to $\Delta y=\frac{1}{2}gt^{2}$.

Step2: Solve for time $t$

First, multiply both sides of the equation $\Delta y=\frac{1}{2}gt^{2}$ by $2$ to get $2\Delta y = gt^{2}$. Then, divide both sides by $g$: $t^{2}=\frac{2\Delta y}{g}$. Finally, take the square - root of both sides. Since time $t>0$ in this context, $t = \sqrt{\frac{2\Delta y}{g}}$.

Answer:

$\sqrt{\frac{2\Delta y}{g}}$