Projectile motion — horizontal - vertical independence, full derivation: Exercises

This page is a self-test ladder. Each rung is harder than the last. Try each problem with the solution hidden, then open the [!recall]- callout to check every step. All symbols come from the parent note the main derivation — nothing new is assumed.

Quick symbol reminder so line one is readable:

Level 1 — Recognition

Can you pick the right formula and plug in?

L1·Q1 — Read off the components

A ball leaves the ground at $u = 50 m/s$ , $θ = 3 7^{\circ}$ (use $cos 3 7^{\circ} = 0.8$ , $sin 3 7^{\circ} = 0.6$ ). Write $u_{x}$ and $u_{y}$ .

The picture below is the object of the whole problem: the launch velocity is one arrow, and its two shadows are the pieces we compute.

Figure — Projectile motion — horizontal - vertical independence, full derivation

Recall Solution

WHAT: split the launch arrow into its two shadows. WHY: the two axes evolve independently, so we need each piece separately. $u_{x} = u cos θ = 50 \times 0.8 = 40 m/s$ $u_{y} = u sin θ = 50 \times 0.6 = 30 m/s$ Look at figure s01 above: the burnt-orange arrow is $u$ , the teal arrow is $u_{x}$ (its floor-shadow), the plum arrow is $u_{y}$ (its wall-shadow). The right triangle they form is exactly what $cos$ (adjacent/hypotenuse) and $sin$ (opposite/hypotenuse) measure.

L1·Q2 — Time of flight

Same ball ( $u_{y} = 30 m/s$ ). Level ground. Find $T$ .

Recall Solution

WHY only $u_{y}$ ? Only the vertical motion decides when $y$ returns to $0$ ; the horizontal glide is irrelevant to timing. $T = \frac{2 u s i n θ}{g} = \frac{2 \times 30}{10} = 6 s$

L1·Q3 — Max height and range

Same ball. Find $H$ and $R$ .

Recall Solution

WHY $H = u_{y}^{2} / (2 g)$ ? In the formula $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ the numerator is $(u sin θ)^{2}$ . But $u sin θ$ is the vertical shadow $u_{y}$ we defined above. So $u^{2} sin^{2} θ = u_{y}^{2}$ , and we may write $H = \frac{u _{y}^{2}}{2 g}$ — using the number $u_{y} = 30$ we already have. $H = \frac{u ^{2} s i n ^{2} θ}{2 g} = \frac{u _{y}^{2}}{2 g} = \frac{3 0 ^{2}}{2 \times 10} = \frac{900}{20} = 45 m$ $R = u_{x} \cdot T = 40 \times 6 = 240 m$ (Check via the range formula: $R = \frac{u ^{2} sin 2 θ}{g} = \frac{2500 \times sin 7 4 ^{\circ}}{10} = \frac{2500 \times 0.9613}{10} \approx 240 m$ ✓.)

Level 2 — Application

Chain two or three steps together.

L2·Q1 — Velocity at a given instant

A projectile is launched at $u = 20 m/s$ , $θ = 6 0^{\circ}$ ( $sin 6 0^{\circ} = 0.866$ , $cos 6 0^{\circ} = 0.5$ ). Find the speed and direction of the velocity at $t = 1 s$ .

Recall Solution

WHAT: find each velocity component at $t = 1$ , then recombine. $u_{x} = 20 \times 0.5 = 10 m/s$ (constant — no horizontal force). $u_{y} = 20 \times 0.866 = 17.32 m/s$ . At $t = 1 s$ : $v_{x} = u_{x} = 10 m/s, v_{y} = u_{y} - g t = 17.32 - 10 \times 1 = 7.32 m/s$ Speed = magnitude of the arrow (Pythagoras on the two shadows): $v = v_{x}^{2} + v_{y}^{2} = 1 0^{2} + 7.3 2^{2} = 100 + 53.6 = 153.6 = 12.39 m/s$ Direction above horizontal (the angle whose tan is rise/run): $ϕ = arctan (\frac{v _{y}}{v _{x}}) = arctan (\frac{7.32}{10}) = 36. 2^{\circ}$ Still climbing (because $v_{y} > 0$ ), just less steeply than launch.

L2·Q2 — Horizontal launch off a cliff

A stone is thrown horizontally at $u = 15 m/s$ from a cliff $80 m$ tall. Find the flight time and how far from the base it lands.

The figure below shows why the timing is a pure free-fall problem: strip away the sideways glide and the vertical drop is identical to a stone released from rest.

Recall Solution

WHY $θ = 0$ ? "Horizontally" means the launch arrow lies flat: $u_{x} = 15$ , $u_{y} = 0$ . Choosing a sign convention. In the master equation, up is positive and the drop is written as $y = u_{y} t - \frac{1}{2} g t^{2}$ ; landing $80 m$ below means $y = - 80$ , giving $- 80 = - 5 t^{2}$ . It is cleaner here to measure distance downward as positive instead: then the fallen distance $d$ grows as $d = \frac{1}{2} g t^{2}$ (the minus sign flips to a plus because "down" now counts as the positive direction, and gravity points that way). Both conventions give the same $t$ ; we pick "down positive" only to avoid carrying a minus sign. Starting from rest vertically ( $u_{y} = 0$ ): $80 = \frac{1}{2} g t^{2} = \frac{1}{2} (10) t^{2} = 5 t^{2} \Rightarrow t^{2} = 16 \Rightarrow t = 4 s$ Horizontal: constant speed × time $x = u_{x} t = 15 \times 4 = 60 m$ Figure s02 above shows the arc: the vertical drop is identical to a free-fall dashed line dropped at the same instant — they reach the ground together.

L2·Q3 — Same range, two angles

Show that $θ = 2 0^{\circ}$ and $θ = 7 0^{\circ}$ give the same range for the same speed, and check for $u = 30 m/s$ .

Recall Solution

WHY they match: $R = \frac{u ^{2} sin 2 θ}{g}$ . For $2 0^{\circ}$ : $sin 4 0^{\circ}$ . For $7 0^{\circ}$ : $sin 14 0^{\circ} = sin (18 0^{\circ} - 4 0^{\circ}) = sin 4 0^{\circ}$ . Same value ⇒ same range. $R = \frac{3 0 ^{2} \times s i n 4 0 ^{\circ}}{10} = \frac{900 \times 0.6428}{10} = 57.85 m (both angles)$ But their times differ: $sin 7 0^{\circ} > sin 2 0^{\circ}$ , so the $7 0^{\circ}$ ball stays up longer and flies higher.

Level 3 — Analysis

Reverse the logic, or combine independence with a twist.

L3·Q1 — Find the launch angle from the flight data

A projectile launched at $u = 40 m/s$ has a time of flight $T = 4 s$ on level ground. Find $θ$ and the range.

Recall Solution

WHAT: invert the time-of-flight formula to solve for the angle. $T = \frac{2 u s i n θ}{g} \Rightarrow sin θ = \frac{g T}{2 u} = \frac{10 \times 4}{2 \times 40} = \frac{40}{80} = 0.5$ WHY $arcsin$ ? It answers "which angle has this sine?" — the inverse that undoes $sin$ . $θ = arcsin (0.5) = 3 0^{\circ}$ Range: $R = u cos θ \cdot T = 40 cos 3 0^{\circ} \times 4 = 40 \times 0.866 \times 4 = 138.6 m$

L3·Q2 — Two stones, race to the ground

From a $45 m$ cliff, stone A is dropped and stone B is thrown horizontally at $12 m/s$ , both at the same instant. Which lands first, and what is the horizontal gap between their landing points?

Recall Solution

The key insight (independence): the horizontal throw of B does not touch its vertical motion. Both stones share the identical vertical equation $45 = \frac{1}{2} g t^{2}$ . $45 = 5 t^{2} \Rightarrow t^{2} = 9 \Rightarrow t = 3 s for BOTH$ So they land simultaneously — neither wins. Stone A lands at the cliff base ( $x_{A} = 0$ ). Stone B travels $x_{B} = 12 \times 3 = 36 m$ . Gap: $Δ x = 36 - 0 = 36 m$

L3·Q3 — Height when velocity is known

A ball launched at $u = 25 m/s$ , $θ = 5 3^{\circ}$ ( $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ). At what height is its speed equal to $20 m/s$ on the way up?

Recall Solution

WHAT: $v_{x}$ is fixed, so the speed condition pins down $v_{y}$ ; then find the height at that $v_{y}$ . $u_{x} = 25 \times 0.6 = 15 m/s$ (constant), $u_{y} = 25 \times 0.8 = 20 m/s$ . Speed $v = 20$ : $v_{x}^{2} + v_{y}^{2} = 2 0^{2} \Rightarrow 1 5^{2} + v_{y}^{2} = 400 \Rightarrow v_{y}^{2} = 400 - 225 = 175$ . Height using vertical energy relation $v_{y}^{2} = u_{y}^{2} - 2 g y$ (from 1-D kinematics): $y = \frac{u _{y}^{2} - v _{y}^{2}}{2 g} = \frac{2 0 ^{2} - 175}{2 \times 10} = \frac{400 - 175}{20} = \frac{225}{20} = 11.25 m$

Level 4 — Synthesis

Build the answer from the parametric equations directly.

L4·Q1 — Cliff launch at an angle

A projectile is fired from the top of a $60 m$ tower at $u = 25 m/s$ , $θ = 3 7^{\circ}$ ( $sin 3 7^{\circ} = 0.6, cos 3 7^{\circ} = 0.8$ ). Find the total flight time and the horizontal distance from the tower base where it lands.

The figure below shows the full arc dropping below the launch line — this is why we cannot use the level-ground shortcut.

Recall Solution

WHAT & WHY: the landing is $60 m$ below launch, so we set $y = - 60$ in the parametric equation (up positive, origin at the launch point). $u_{x} = 25 \times 0.8 = 20 m/s$ , $u_{y} = 25 \times 0.6 = 15 m/s$ . $y = u_{y} t - \frac{1}{2} g t^{2} \Rightarrow - 60 = 15 t - 5 t^{2}$ Rearrange into a standard quadratic: $5 t^{2} - 15 t - 60 = 0 \Rightarrow t^{2} - 3 t - 12 = 0$ WHY the quadratic formula? two unknown times satisfy "height $= - 60$ ": one physical (positive), one unphysical (negative, before launch). We keep the positive root. $t = \frac{3 + 9 + 48}{2} = \frac{3 + 57}{2} = \frac{3 + 7.550}{2} = 5.275 s$ $x = u_{x} t = 20 \times 5.275 = 105.5 m$ Figure s03 above overlays this cliff arc; the horizontal reach is much larger than the level-ground formula would give.

L4·Q2 — Clearing a wall

A ball is kicked at $u = 20 m/s$ , $θ = 4 5^{\circ}$ . A wall $3 m$ tall stands $30 m$ away. Does the ball clear it? (Use $sin 4 5^{\circ} = cos 4 5^{\circ} = 0.707$ .)

Recall Solution

WHY the trajectory equation? We are asked for the height $y$ at a given horizontal position $x = 30$ , not at a given time. The trajectory equation gives $y$ directly in terms of $x$ , so we never need $t$ . It comes straight from the parametric pair: from $x = u cos θ t$ solve $t = \frac{x}{u cos θ}$ , then substitute into $y = u sin θ t - \frac{1}{2} g t^{2}$ . The first term becomes $u sin θ \cdot \frac{x}{u cos θ} = x tan θ$ , and the second becomes $\frac{1}{2} g \cdot \frac{x ^{2}}{u ^{2} cos ^{2} θ}$ , giving: $y = x tan θ - \frac{g x ^{2}}{2 u ^{2} c o s ^{2} θ}$ Now plug in $tan 4 5^{\circ} = 1$ , $cos^{2} 4 5^{\circ} = 0.5$ , $u^{2} = 400$ : $y = 30 (1) - \frac{10 \times 3 0 ^{2}}{2 \times 400 \times 0.5} = 30 - \frac{9000}{400} = 30 - 22.5 = 7.5 m$ The ball is at $7.5 m$ when it reaches the wall, which is above $3 m$ ⇒ it clears the wall by $4.5 m$ .

Level 5 — Mastery

Prove a relationship or optimise.

L5·Q1 — Prove the range–height link

Show that for level ground, $R = \frac{4 H}{tan θ}$ .

Recall Solution

WHAT: express both $R$ and $H$ with the same symbols and divide. $R = \frac{u ^{2} s i n 2 θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g}, H = \frac{u ^{2} s i n ^{2} θ}{2 g}$ Form the ratio (the $u^{2} / g$ cancels): $\frac{R}{H} = \frac{2 s i n θ c o s θ}{\frac{1}{2} s i n ^{2} θ} = \frac{4 c o s θ}{s i n θ} = \frac{4}{t a n θ}$ Therefore $R = \frac{4 H}{tan θ}$ . Sanity check at $θ = 4 5^{\circ}$ : $tan 4 5^{\circ} = 1 \Rightarrow R = 4 H$ , i.e. the arc is four times wider than it is tall at the max-range angle.

L5·Q2 — Derive $T$ and $H$ by calculus (no memorised formula)

Starting only from $a_{y} = - g$ and the launch data $u = 30 m/s$ , $θ = 3 0^{\circ}$ , use integration to obtain $v_{y} (t)$ , $y (t)$ , then find $T$ and $H$ from scratch.

Recall Solution

Step 1 — integrate acceleration to velocity. WHY integrate? $a_{y} = d v_{y} / d t$ , so undoing the derivative recovers velocity: $v_{y} (t) = \int (- g) d t = - g t + C .$ At $t = 0$ , $v_{y} = u sin θ = 30 \times 0.5 = 15$ , so $C = 15$ . Thus $v_{y} = 15 - 10 t$ . Step 2 — integrate velocity to position. $y (t) = \int (15 - 10 t) d t = 15 t - 5 t^{2} (origin at launch, C = 0) .$ Step 3 — time of flight: land when $y = 0$ : $0 = t (15 - 5 t) \Rightarrow t = 0 or t = 3 s \Rightarrow T = 3 s .$ Step 4 — max height: top when $v_{y} = 0$ : $15 - 10 t = 0 \Rightarrow t = 1.5 s$ . Plug in: $H = 15 (1.5) - 5 (1.5)^{2} = 22.5 - 11.25 = 11.25 m .$ These match $T = \frac{2 u s i n θ}{g} = \frac{2 \times 15}{10} = 3$ and $H = \frac{u _{y}^{2}}{2 g} = \frac{225}{20} = 11.25$ . ✓

L5·Q3 — Optimal angle for a cliff-edge shot (challenge)

From ground level you want maximum range at $u = 20 m/s$ . Confirm $θ = 4 5^{\circ}$ is optimal, then state the maximum range.

Recall Solution

WHY $4 5^{\circ}$ : $R = \frac{u ^{2} sin 2 θ}{g}$ is largest when $sin 2 θ$ is largest. The sine function maxes at $1$ when its argument is $9 0^{\circ}$ , so $2 θ = 9 0^{\circ} \Rightarrow θ = 4 5^{\circ}$ . $R_{m a x} = \frac{u ^{2}}{g} = \frac{2 0 ^{2}}{10} = 40 m .$ Any deviation (say $3 0^{\circ}$ : $sin 6 0^{\circ} = 0.866$ , giving $R = 34.6 m$ ) falls short — confirming the peak.

Recall One-line answer key (numbers only)

L1: $u_{x} = 40, u_{y} = 30$ ; $T = 6 s$ ; $H = 45 m, R = 240 m$ . L2: $v = 12.39 m/s$ at $36. 2^{\circ}$ ; $t = 4 s, x = 60 m$ ; $R = 57.85 m$ . L3: $θ = 3 0^{\circ}, R = 138.6 m$ ; both land in $3 s$ , gap $36 m$ ; $y = 11.25 m$ . L4: $t = 5.275 s, x = 105.5 m$ ; height at wall $= 7.5 m$ (clears). L5: $R = 4 H / tan θ$ ; $T = 3 s, H = 11.25 m$ ; $R_{m a x} = 40 m$ .