Projectile Motion: Edge Cases and Subtle Traps (1)

Question

A projectile is launched with speed $u = 20\text{ m/s}$ at an angle of $60°$ above the horizontal from ground level. (a) Find the time of flight. (b) Find the maximum height. (c) Find the horizontal range. (d) At what other angle would the same range be obtained? Take $g = 10\text{ m/s}^2$.

Solution — Step by Step

Why This Works

Projectile motion decouples into independent horizontal (uniform velocity) and vertical (uniform acceleration $-g$) motions. That decoupling is the only physics in the chapter — everything else is kinematics applied separately to the two axes.

The complementary-angle result comes from $R = u^2 \sin(2\theta)/g$. Since $\sin(2\theta) = \sin(180° - 2\theta) = \sin(2(90° - \theta))$, both $\theta$ and $90° - \theta$ produce the same range.

Alternative Method

Using the trajectory equation $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$, set $y = 0$ for landing — this gives $x = R$ directly. Faster when only the range is asked.

Common Mistake