Quick questions on Projectile motion: parametric equations, range, maximum height and time of flight

The figures below build a launch from the ground one feature at a time, in the order you should set the problem up.

Eliminate $t$ from the position equations. From $x = V \cos \alpha \cdot t$, $t = \frac{x}{V \cos \alpha}$. Substitute into $y$:

If the projectile starts at height $h$, then $y(t) = h + V \sin \alpha \cdot t - \tfrac{1}{2} g t^2$. The time of flight is now the larger root of this quadratic in $t$, and the symmetric range formula no longer applies - you must solve the quadratic. This is the single most common way the standard formulas are made to fail, so check the launch height before reaching for $R = \frac{V^2\sin 2\alpha}{g}$.

Magnitude of velocity at time $t$:

For ground inclined at angle $\beta$ to the horizontal, the projectile lands when its path meets the line of the slope, so you set $y(t) = \pm x(t)\tan\beta$ (sign depending on uphill or downhill) and solve for $t$. This is a harder Section II application; the method is unchanged, only the landing condition is.

$R = V \cos \alpha \cdot T$.

set $\dot y(t) = 0$, giving $t = \frac{V \sin \alpha}{g}$, then substitute.

The range formula uses $\sin 2 \alpha$, not $\sin^2 \alpha$. The double angle comes from $2\sin \alpha \cos \alpha = \sin 2 \alpha$.