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?

Explanation:

Step1: Identify the relevant kinematic equation

We use the equation $h = v_0t+\frac{1}{2}gt^2$. Since the ball is dropped ($v_0 = 0$), the equation simplifies to $h=\frac{1}{2}gt^2$, where $h = 38.0m$ and $g = 9.8m/s^2$.

Step2: Solve for time $t$

Rearranging $h=\frac{1}{2}gt^2$ for $t$, we get $t=\sqrt{\frac{2h}{g}}$. Substituting $h = 38.0m$ and $g=9.8m/s^2$ into the formula: $t=\sqrt{\frac{2\times38.0}{9.8}}\approx 2.79s$.

Step3: Find the final - speed

We use the equation $v = v_0+gt$. Since $v_0 = 0$, then $v = gt$. Substituting $g = 9.8m/s^2$ and $t = 2.79s$, we get $v=9.8\times2.79 = 27.3m/s$.

Answer:

The ball remained in the air for approximately $2.79s$ and its speed just before striking the ground was approximately $27.3m/s$.