Answer:
Explanation:
a) To find the initial velocity, $V_0$ of the bullet, we can use the formula for maximum height,$h$ attained by an object when it's thrown straight up into the air.$$\begin{aligned} h &= \frac{V_0^2}{2g} \\ V_0^2 &= 2gh \\ V_0 &= \sqrt{2gh} \end{aligned}$$where $g$ is the acceleration due to gravity. We can plug in the value of $h=946 \mathrm{~m}$ and $g=9.8 \mathrm{~m/s^2}$ and solve for $V_0$.$$ V_0 = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \mathrm{~m/s^2} \cdot 946 \mathrm{~m}} \approx \boxed{437.0 \mathrm{~m/s}}$$Therefore, the initial velocity of the bullet was approximately $437.0 \mathrm{~m/s}$.
b) To find the time of flight, $t$ for when the bullet travels up and then returns to the ground, we can use the formula for the time of flight,$t$.$$t = \frac{2V_0}{g}$$where $g$ is the acceleration due to gravity. We can plug in the value of $V_0=437.0 \mathrm{~m/s}$ and $g=9.8 \mathrm{~m/s^2}$ and solve for $t$.$$ t = \frac{2V_0}{g} = \frac{2\cdot437.0 \mathrm{~m/s}}{9.8 \mathrm{~m/s^2}} \approx \boxed{89.0 \mathrm{~s}}$$Therefore, the time of flight for the bullet was approximately $89.0 \mathrm{~s}$.