a)  What is the flight time of the cannonball?

$t=5\ \mathrm{s}$

a) For parabolic trajectories, the total flight time can be determined by doubling the amount of time it takes for the projectile to reach its apex. However, this trajectory is NOT parabolic. As the initial position is $25\ \mathrm{m}$, the final position is $0\ \mathrm{m}$, the initial velocity is

$$v_0=v_0\sin{\theta }$$

$$v_0=40\sin{30^\circ }$$

$$v_0=40\left(\frac{1}{2}\right)$$

$$v_0=20\ \mathrm{\displaystyle \frac{m}{s}}$$

and the acceleration is $−10\ \mathrm{\displaystyle \frac{m}{\ s^2}}$, the missing variable is the final velocity so the flight time of the cannonball can be computed using Big Five #3:

$$x_y=x_0+v_0t+\frac{1}{2}at^2$$

$$0=25\ \mathrm{m+20\ \frac{m}{s}}\cdot t+\frac{1}{2}\left(-10\ \mathrm{\displaystyle \frac{m}{\ s^2}}\right)\cdot t^2$$

$$0=25\ \mathrm{m+20\ \frac{m}{s}}\cdot t-5\ \mathrm{\displaystyle \frac{m}{\ s^2}}\cdot t^2$$

$$0=-5\ \mathrm{\displaystyle \frac{m}{\ s^2}}\cdot t^2+20\ \mathrm{\frac{m}{s}}\cdot t+25\ \mathrm{m}$$

Applying the quadratic formula, find that:

$$t=\frac{-20\pm \sqrt{20^2-4(25)(-5)}}{2(-5)}$$

$$t=\frac{-20\pm \sqrt{400+500}}{-10}$$

$$t=\frac{-20\pm \sqrt{900}}{-10}$$

$$t=\frac{-20\pm 30}{-10}$$

$$t=-1\ \mathrm{s}\ \ \mathrm{or}\ \ t=5\ \mathrm{s}$$

As time cannot be a negative value, the flight time of the cannonball is $5\ \mathrm{seconds}$.

b)  What is the range of the cannonball?

$\Delta x=173\ \mathrm{m}$

b) As the horizontal speed of a projectile is constant, the range is given by:

$$\Delta x=v_{0x}t$$

$$\Delta x=40\ \mathrm{\frac{m}{s}}\cdot \cos{30^\circ }\cdot 5\ \mathrm{s}$$

$$\Delta x=173\ \mathrm{m}$$

c)  In the two graphs below, plot the horizontal speed of the projectile versus time and the vertical speed of the projectile versus time (from the initial launch of the projectile to the instant it strikes the ground).

c) The cannonball is fired with an initial horizontal velocity of:

$$v_{0x}=40\ \mathrm{\frac{m}{s}}\cdot \cos{30^\circ }$$

$$v_{0x}=34.6\ \mathrm{\displaystyle \frac{m}{s}}$$

As there is no horizontal acceleration the horizontal speed of the projectile remains constant throughout the entire flight, leading to a flat line with a $y$‑value of $34.6\ \mathrm{\displaystyle \frac{m}{s}}$ over the flight time of $5\ \mathrm{s}$. The horizontal speed vs. time graph can then be plotted

The initial vertical velocity of the cannonball is:

$$v_{0x}=40\ \mathrm{\frac{m}{s}}\cdot \sin{30^\circ }$$

$$v_{0x}=20\ \mathrm{\frac{m}{s}}$$

In the vertical direction, there is a constant acceleration due to gravity $(a = 10\ \mathrm{\displaystyle \frac{m}{\ s^2}})$. As the slope of the velocity time graph is equal to the acceleration, the resulting velocity time graph is a linear line with a negative slope since acceleration due to gravity points downwards. As the magnitude of the acceleration of gravity is $10\ \mathrm{\displaystyle \frac{m}{\ s^2}}$, the vertical velocity of the cannonball thus decreases by $10\ \mathrm{\displaystyle \frac{m}{s}}$ every second. With an initial velocity of $20\ \mathrm{\displaystyle \frac{m}{s}}$, this means that after $2\ \mathrm{s}$, the vertical velocity of the projectile is $0\ \mathrm{\displaystyle \frac{m}{s}}$. After a total of $5\ \mathrm{s}$, the vertical velocity of the projectile is $-30\ \mathrm{\displaystyle \frac{m}{s}}$. This can be visualized in the vertical velocity vs. time graph below:

As the problem asks for the graph of the vertical speed vs. time, the absolute value of the graph must be drawn to get the correct plot of