a student drops a ball from a height of 49.0m. determine all unknowns and answer the following questions…

Explanation:

Step1: Find time in air

Use the equation $h = v_0t+\frac{1}{2}gt^2$. Since the ball is dropped, $v_0 = 0$. The equation simplifies to $h=\frac{1}{2}gt^2$. Solving for $t$, we get $t=\sqrt{\frac{2h}{g}}$. Given $h = 49.0m$ and $g = 9.8m/s^2$. $t=\sqrt{\frac{2\times49.0}{9.8}}$

Step2: Calculate time value

$t=\sqrt{10}\approx 3.16s$

Step3: Find speed before hitting ground

Use the equation $v = v_0+gt$. Since $v_0 = 0$, $v = gt$. Substitute $g = 9.8m/s^2$ and $t = 3.16s$. $v=9.8\times3.16$

Step4: Calculate speed value

$v\approx31.0m/s$

Answer:

How long did the ball remain in the air? 3.16 s What was the ball's speed just before striking the ground? 31.0 m/s