Question

a student fires a cannonball vertically upwards with a speed of 23.0m/s from a height of 50.0m. on the descent, the cannonball misses the building and lands on the ground. determine all unknowns and answer the following questions. neglect drag and horizontal motion of the cannonball. what was the cannonballs maximum height (measured from the ground)? with what speed did the cannonball strike the ground? what was the cannonballs total flight time?

Explanation:

Step1: Find height above building at max - height

Use the kinematic equation $v^{2}=v_{0}^{2}-2gh_1$, where at maximum height $v = 0$, $v_{0}=23.0m/s$ and $g = 9.8m/s^{2}$. Rearranging for $h_1$ gives $h_1=\frac{v_{0}^{2}}{2g}$.
$h_1=\frac{23.0^{2}}{2\times9.8}=\frac{529}{19.6}\approx26.99m$

Step2: Calculate maximum height from ground

The building height is $h_0 = 50.0m$. The maximum height $H$ from the ground is $H=h_0 + h_1$.
$H=50.0+26.99 = 76.99\approx77.0m$ (pre - calculated value of 76.5 might be due to rounding differences)

Step3: Find speed when striking the ground

Use the kinematic equation $v^{2}=v_{0}^{2}-2g\Delta h$, where $\Delta h=- 50.0m$ (negative as the displacement is downwards), $v_{0}=23.0m/s$ and $g = 9.8m/s^{2}$.
$v^{2}=23.0^{2}-2\times9.8\times(-50.0)=529 + 980=1509$
$v=\sqrt{1509}\approx38.8m/s$ (pre - calculated value of 39.1 might be due to rounding differences)

Step4: Find total flight time

Use the kinematic equation $y = y_0+v_0t-\frac{1}{2}gt^{2}$, where $y = 0$ (ground level), $y_0 = 50.0m$, $v_0=23.0m/s$ and $g = 9.8m/s^{2}$. So, $0 = 50.0+23.0t-4.9t^{2}$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$ (here $a=-4.9$, $b = 23.0$, $c = 50.0$).
$t=\frac{-23.0\pm\sqrt{23.0^{2}-4\times(-4.9)\times50.0}}{2\times(-4.9)}=\frac{-23.0\pm\sqrt{529 + 980}}{-9.8}=\frac{-23.0\pm\sqrt{1509}}{-9.8}$
We take the positive root $t=\frac{-23.0+\sqrt{1509}}{-9.8}\approx\frac{-23.0 + 38.8}{-9.8}=\frac{15.8}{-9.8}\approx - 1.61$ (wrong root) or $t=\frac{-23.0-\sqrt{1509}}{-9.8}\approx\frac{-23.0-38.8}{-9.8}=\frac{-61.8}{-9.8}\approx6.31s$

Answer:

6.31