Question

a student drops a ball from a height of 38.0m. determine all unknowns and answer the following questions. neglect drag. how long did the ball remain in the air? what was the balls speed just before striking the ground? unit unit

Explanation:

Step1: Identify the relevant kinematic - equation for speed

We use the equation $v^{2}=v_{0}^{2}+2a\Delta y$. The ball is dropped ($v_{0} = 0$), $a = g=9.8\ m/s^{2}$ and $\Delta y=38.0\ m$.
$$v^{2}=0 + 2gh$$

Step2: Solve for the speed $v$

Substitute $g = 9.8\ m/s^{2}$ and $h = 38.0\ m$ into the equation.
$v=\sqrt{2gh}=\sqrt{2\times9.8\times38.0}\approx\sqrt{744.8}\approx27.3\ m/s$

Step3: Identify the relevant kinematic - equation for time

We use the equation $\Delta y=v_{0}t+\frac{1}{2}at^{2}$. Since $v_{0} = 0$, the equation simplifies to $\Delta y=\frac{1}{2}at^{2}$.
$\Delta y=\frac{1}{2}gt^{2}$

Step4: Solve for the time $t$

Rearrange the equation for $t$: $t=\sqrt{\frac{2\Delta y}{g}}$. Substitute $\Delta y = 38.0\ m$ and $g = 9.8\ m/s^{2}$.
$t=\sqrt{\frac{2\times38.0}{9.8}}\approx\sqrt{\frac{76}{9.8}}\approx\sqrt{7.76}\approx2.79\ s$

Answer:

The ball's speed just before striking the ground is approximately $27.3\ m/s$ (units: m/s). The ball remains in the air for approximately $2.79\ s$ (units: s).