a)  Calculate the maximum height, $H$, of the projectile.

$H=\displaystyle \frac{{v_0}^2\sin^2{\theta_{0}}}{2g}$

a) The maximum height of the projectile occurs at the time at which its vertical velocity drops to zero:

$$v_y\overset{\mathrm{set}}{=}0$$

$$v_{0y}-gt=0$$

$$t=\frac{v_{0y}}{g}$$

The vertical displacement of the projectile at this time is computed as follows:

$$\Delta y=v_{0y}t-\frac{1}{2}gt^2$$

$$H=v_{0y}\frac{v_{0y}}{g}-\frac{1}{2}g\left(\frac{v_{0y}}{g}\right)^2$$

$$H=\frac{{v_0}^2y}{2g}$$

$$H=\frac{{v_0}^2\sin^2{\theta_{0}}}{2g}$$

b)  Calculate the (horizontal) range, $R$, of the projectile.

$R=\displaystyle \frac{2{v_0}^2\sin{\theta_0}\cos{\theta_0}}{g}\ \ \mathrm{or}\ \ \frac{{v_0}^2\sin{2\theta_0}}{g}$

b) The total flight time is equal to twice the time computed in part a):

$$t_t=2t=2\frac{v_{0y}}{g}$$

The horizontal displacement at this time gives the projectile’s range:

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

$$R=v_{0x}t_t$$

$$R=\frac{v_{0x}\cdot 2v_{0y}}{g}$$

$$R=\frac{2{v_0}^2\sin{\theta_0}\cos{\theta_0}}{g}\ \ \mathrm{or}\ \ \frac{{v_0}^2\sin{2\theta_0}}{g}$$

c)  For what value of $\theta_0$ will the range be maximized?

$\theta_0=45^\circ$

c) For any given value of $v_0$, the range,

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

$$R=v_{0x}t_t$$

$$R=\frac{v_{0x}\cdot 2v_{0y}}{g}$$

$$R=\frac{2{v_0}^2\sin{\theta_0}\cos{\theta_0}}{g}\ \ \mathrm{or}\ \ \frac{{v_0}^2\sin{2\theta_0}}{g}$$

will be maximized when $\sin{2\theta_0}$ is maximized. This occurs when $\sin{2\theta_0}=90^\circ$, that is, when $\theta_0=45^\circ$.

d)  If $0 < h < H$, compute the time that elapses between passing through the horizontal line of height h in both directions (ascending and descending); that is, compute the time required for the projectile to pass through the two points shown in this figure:

$\Delta t=\displaystyle \frac{2\sqrt{v^2_{0y}-2gh}}{g}$

d) Set the general expression for the projectile’s vertical displacement equal to h and solve for the two values of t $\bigg($assuming that $g=+10\ \mathrm{\displaystyle \frac{m}{\ s^2}}\bigg):$

$$v_{0y}t-\frac{1}{2}gt^2\overset{\mathrm{set}}{=}h$$

Applying the quadratic formula, find that

$$t=\frac{v_{0y}\pm \sqrt{\left(-{v_{0y}}\right)^2-4\left(\displaystyle \frac{1}{2}g\right)(h)}}{2\left(\displaystyle \frac{1}{2}g\right)}$$

$$t=\frac{v_{0y}\pm \sqrt{{v^2_{0y}}-2gh}}{g}$$

Therefore, the two times at which the projectile crosses the horizontal line at height h are

$$t_1=\frac{v_{0y}-\sqrt{v^2_{0y}-2gh}}{g}$$

and

$$t_2=\frac{v_{0y}+\sqrt{v^2_{0y}-2gh}}{g}$$

so the amount of time that elapses between these events is

$$\Delta t=t_2-t_1$$

$$\Delta t=\frac{2\sqrt{v^2_{0y}-2gh}}{g}$$