Exercises — Range, max height, time of flight — all derived

WHAT is asked: time in the air $\Rightarrow$ time of flight $T=\frac{2u\sin \theta }{g}$. WHY this one: "in the air" = from launch until it lands. That is exactly what $T$ measures. Not $R$ (that's distance), not $H$ (that's height). No number crunch needed — the answer is recognizing $T=\frac{2u\sin \theta }{g}$.

Look at how $u$ appears:

• $R=\frac{u^2\sin 2\theta }{g}$ — $u$ is squared. Doubling $u$ gives $2^2=4\times$ the range.
• $T=\frac{2u\sin \theta }{g}$ — $u$ appears to the first power. Doubling $u$ gives $2\times$ the time. WHY it matters: reading the exponent of a variable is the physics. Range grows faster than time because doubling speed helps you both fly longer and move forward faster — those two effects multiply.

Time: $T=\frac{2(30)(0.707)}{10}=\frac{42.4}{10}=4.24\text{s}$. Height: $H=\frac{(30{)}^2(0.707{)}^2}{2(10)}=\frac{900(0.5)}{20}=\frac{450}{20}=22.5\text{m}$. Why $(0.707{)}^2=0.5$? Because ${\sin }^{2}45^{\circ }=\frac{1}{2}$ exactly. Range: $R=\frac{(30{)}^2\sin 90^{\circ }}{10}=\frac{900(1)}{10}=90\text{m}$. Sanity: at $45^{\circ }$, $R=u^2/g=900/10=90$. ✓ This is the maximum possible range for this speed.

Start from $R=\frac{u^2\sin 2\theta }{g}$: $20=\frac{(20{)}^2\sin 2\theta }{10}=\frac{400\sin 2\theta }{10}=40\sin 2\theta \Rightarrow \sin 2\theta =\mathrm{0.5.}$ WHY two answers: the equation $\sin 2\theta =0.5$ has two solutions in the valid range $0^{\circ }<2\theta <180^{\circ }$: $2\theta =30^{\circ }\Rightarrow \theta =15^{\circ },2\theta =150^{\circ }\Rightarrow \theta =75^{\circ }.$ These are complementary ($15^{\circ }+75^{\circ }=90^{\circ }$) — the low flat shot and the high lob that land in the same spot. See the twin trajectories in the figure below.

(a) Ranges. $R_{20}=\frac{625\sin 40^{\circ }}{10}$ and $R_{70}=\frac{625\sin 140^{\circ }}{10}$. Since $\sin 140^{\circ }=\sin (180^{\circ }-140^{\circ })=\sin 40^{\circ }$, both equal $\frac{625(0.643)}{10}\approx 40.2\text{m}$. Equal. ✓ Why: $20^{\circ }$ and $70^{\circ }$ are complementary, and $\sin 2\theta =\sin (180^{\circ }-2\theta )$ makes their ranges identical.

(b) Heights. $H\propto {\sin }^2\theta$, so $\frac{H_{70}}{H_{20}}=\frac{{\sin }^{2}70^{\circ }}{{\sin }^{2}20^{\circ }}={\left(\frac{0.940}{0.342}\right)}^2\approx (2.749{)}^2\approx \mathrm{7.55.}$ The steeper shot climbs about 7.5× higher while landing in the same spot. Equal range does not mean equal path.

Key idea: find $u\sin \theta$ first, because both $T$ and $H$ depend only on the vertical component $u_y=u\sin \theta$. From $T=\frac{2u\sin \theta }{g}$: $u\sin \theta =\frac{gT}{2}=\frac{10\cdot 4}{2}=20\text{m/s}$. Cross-check with height: $H=\frac{(u\sin \theta {)}^2}{2g}=\frac{20^2}{2(10)}=\frac{400}{20}=20\text{m}$. ✓ Consistent. So we know $u\sin \theta =20$, but that's one equation with two unknowns ($u$ and $\theta$). $T$ and $H$ alone cannot separate them — every launch with the same vertical component gives the same $T$ and $H$. Conclusion: $u\sin \theta =20\text{m/s}$ is fully determined, but $u$ and $\theta$ individually are not — we'd need the range or the horizontal speed too. Recognizing this under-determination is the answer.

Energy setup. At the peak, only the vertical part of the velocity is zero; the horizontal speed $u\cos \theta$ survives. So the kinetic energy that gets "spent" climbing is the vertical part only. Per unit mass, energy conservation between launch and peak: $\frac{1}{2}(u\sin \theta {)}^2=gH\Rightarrow H=\frac{(u\sin \theta {)}^2}{2g}.$ Why only $u\sin \theta$? The horizontal speed never changes and never converts to height, so it cancels from both sides. Numbers: $u\sin \theta =20(0.5)=10\text{m/s}$, so $H=\frac{10^2}{2(10)}=\frac{100}{20}=5\text{m}$. Matches the kinematic formula: $H=\frac{(20{)}^2(0.5{)}^2}{2(10)}=\frac{400(0.25)}{20}=5\text{m}$. ✓ Two completely different routes (energy vs. equations of motion) agree — the mark of a correct result.

Break into two axes (the whole point of projectile motion — see Projectile Motion — components & independence):

• Horizontal velocity stays constant: $v_x=17.32\text{m/s}$.
• Vertical velocity decays under gravity: $v_y=u_y-gt=10-10(1)=0\text{m/s}$. Interpretation: $v_y=0$ at $t=1$s means the ball is at its peak at exactly this instant. (Check: $t_{\text{up}}=u\sin \theta /g=10/10=1$s. ✓) Speed = $\sqrt{v_x^2+v_y^2}=\sqrt{17.32^2+0^2}=17.32\text{m/s}$. Direction: $\tan \varphi =v_y/v_x=0/17.32=0\Rightarrow \varphi =0^{\circ }$ — moving perfectly horizontally, as expected at the top of the arc.

Why the standard $R$ formula fails: $R=\frac{u^2\sin 2\theta }{g}$ assumes launch and landing at the same height. Here they differ by $15$m, so we must re-derive from the vertical equation. Take up as positive, origin at launch point. Landing is $15$m below, so $y=-15$. $y=u\sin \theta t-\frac{1}{2}g{t}^2\Rightarrow -15=(20)(0.5)t-\frac{1}{2}(10)t^2=10t-5t^2.$ Rearrange into standard quadratic form: $5t^2-10t-15=0\Rightarrow t^2-2t-3=0\Rightarrow (t-3)(t+1)=0.$ Roots: $t=3\text{s}$ and $t=-1\text{s}$. Discard the negative root (before launch); the physical flight time is $\boxed{t=3\text{s}}$. (b) Horizontal distance: horizontal motion is uniform, so $x=u\cos \theta \cdot t=20(0.866)(3)=51.96\approx 52.0\text{m}.$ Notice this is larger than the level-ground range would be — the extra fall time lets the steady horizontal motion carry it farther.

With $\sin 90^{\circ }=1$, $\cos 90^{\circ }=0$, $\sin (2\cdot 90^{\circ })=\sin 180^{\circ }=0$:

• $T=\frac{2(20)(1)}{10}=4\text{s}$ — up for $2$s, down for $2$s. Sensible.
• $H=\frac{(20{)}^2(1{)}^2}{2(10)}=\frac{400}{20}=20\text{m}$ — the highest possible for this speed, since all the speed is vertical.
• $R=\frac{(20{)}^2\sin 180^{\circ }}{10}=\frac{400(0)}{10}=0\text{m}$. Interpretation of $R=0$: a vertical throw has no horizontal component ($u\cos 90^{\circ }=0$), so it comes straight back down to the launch point — zero horizontal distance. The formula correctly reports this degenerate case.

With $\sin 0^{\circ }=0$: $T=0$, $H=0$, $R=0$. Why all zero and why it's right: the formulas assume launch and landing at the same height (the ground). If you launch horizontally from the ground, the vertical component is zero, so gravity pulls the ball straight into the floor with no airtime — it never gets airborne. $T=H=R=0$ faithfully describes "the ball is already on the ground." Contrast with L5.1: a horizontal launch from a cliff would give a nonzero range — but then you must re-derive $T$ from $y=-h$, because the same-height assumption is broken. The zeros here are a feature, not a bug: they flag that the formula's assumption has hit its boundary.