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

A projectile launched from the origin with initial speed $V$ at angle $\alpha$ above the horizontal, with gravitational acceleration $g$ downward and no air resistance:

Time of flight (on horizontal ground): set $y(t) = 0$.

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 $t$; the range is no longer the symmetric formula.

Magnitude of velocity at time $t$:

For ground inclined at angle $\beta$ below horizontal at launch (downhill) or above (uphill), the time of flight and range require setting $y(t) = -x(t) \tan \beta$ (for downhill) or solving the relevant intersection. This is a more involved application and shows up in harder Section II questions.

Same launch. $H = \frac{V^2 \sin^2 \alpha}{2 g} = \frac{400 \cdot 0.25}{19.6} \approx 5.10$ m.

The ball reaches what height at $t = 1.5$ s?

Find the Cartesian trajectory for a ball at $25$ m/s at $40^\circ$.

A ball is thrown at $V = 15$ m/s at $\alpha = 60^\circ$. Find the speed at landing.

A footballer kicks the ball at fixed speed. What launch angle gives maximum range?

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

$V \cos \alpha$ is the horizontal component, $V \sin \alpha$ is the vertical. Mixing them swaps the trajectory.

The range formula uses $\sin 2 \alpha$, not $\sin^2 \alpha$. The double angle is from the product $\sin \alpha \cos \alpha = \frac{1}{2} \sin 2 \alpha$.

Many HSC questions launch from a cliff or table. The position equation needs the offset $h$ for $y(0)$.