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 kinematic equation for vertical - motion

The ball is in free - fall, and the initial velocity $v_0 = 0\ m/s$. The displacement $y - y_0=- 38.0\ m$ (taking downwards as negative and the initial position $y_0 = 0$ and the final position $y=-38.0\ m$), and the acceleration $a=-g=- 9.8\ m/s^{2}$. We use the equation $y - y_0=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$, the equation simplifies to $y - y_0=\frac{1}{2}at^{2}$.
$y - y_0=\frac{1}{2}at^{2}$

Step2: Solve for the time $t$

Substitute $y - y_0=-38.0\ m$ and $a = - 9.8\ m/s^{2}$ into the equation $y - y_0=\frac{1}{2}at^{2}$.
$-38.0=\frac{1}{2}\times(-9.8)t^{2}$
$t^{2}=\frac{-38.0\times2}{-9.8}=\frac{76}{9.8}$
$t=\sqrt{\frac{76}{9.8}}\approx2.78\ s$

Step3: Identify the kinematic equation for final velocity

We use the equation $v = v_0+at$. Since $v_0 = 0\ m/s$, $a=-9.8\ m/s^{2}$ and $t = 2.78\ s$.
$v = v_0+at$

Step4: Solve for the final velocity $v$

Substitute $v_0 = 0\ m/s$, $a=-9.8\ m/s^{2}$ and $t = 2.78\ s$ into the equation $v = v_0+at$.
$v=0+(-9.8)\times2.78=- 27.24\ m/s$ (the negative sign indicates the direction is downwards, and speed is the magnitude, so the speed is $27.24\ m/s$)

Answer:

How long did the ball remain in the air? $2.78\ s$
What was the ball's speed just before striking the ground? $27.24\ m/s$