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In the flat-surface, constant-gravitational acceleration model, a coasting projectile conserves horizontal velocity. If it arrives at the top of its trajectory with horizontal speed $v_o$, its earlier velocity at angle $\theta$ above the horizontal has vertical component $v_o\tan \theta$. Its remaining height gain is therefore $v_o^2{\tan }^2\theta /(2g)$. The displayed burnout height follows. This is a local projectile motion approximation, not an exact Kepler orbit. Computing $v_o=\sqrt{g(R_E+h_o)}$ assumes constant gravitational magnitude also at the orbit; computing $v_o=\sqrt{g_0{R}_E^2/(R_E+h_o)}$ instead assumes an inverse-square law with surface value $g_0$. Those assumptions must be distinguished.

The description of motion through position, velocity and acceleration, before specifying the forces responsible. For constant acceleration these are related by integrating time derivatives. Projectile motion under an approximately uniform gravitational acceleration is an elementary example.

Particle dynamics predicts the motion of bodies whose internal size and rotation can be neglected for the question being studied. A point particle has a position and mass; applied forces determine its acceleration through Newton's second law. For fixed mass in an inertial frame, the basic problem iswith an initial position and velocity.

A ball's projectile motion can be approximated this way when its size is small compared with its path and spin has negligible influence. Air drag introduces a velocity-dependent force; contact with a surface may introduce mechanical constraints. The same method extends to interacting particles, where momentum conservation and conservation of energy can simplify the coupled equations of motion.

Isaac Newton made the greater contribution if the criterion is the scope and explanatory integration of mathematical natural philosophy; Galileo Galilei has the stronger claim if the criterion is the formative transformation of how terrestrial motion could be investigated. The first criterion gives a defensible overall verdict for Isaac Newton, but a serious comparison must show what each changed rather than award credit according to who worked later.

Galileo Galilei's contribution was not simply to attach numbers to familiar events. In the Third Day of Two New Sciences by Galileo Galilei, he defines uniformly accelerated motion by equal additions of speed in equal intervals of time, then derives consequences. Starting from rest, the modern notation is $v=at$; the mean speed over the interval is $at/2$, soHence distances in successive equal intervals have the ratios $1:3:5:7:\dots$. The novelty lies in making a precisely specified kind of motion the subject of mathematical demonstration, with observed falls assessed against its consequences. This is not the same as deriving the underlying physical cause of gravity.^[ref]

The reported apparatus shows how such an idealization could be investigated. A smooth bronze ball descended in a straight grooved wooden channel, raised at one end. Water discharged through a small opening into a vessel during the descent; its collected mass served as a measure of elapsed time. Repeating runs over different portions of the channel compared the predicted squared dependence of time on distance. The inclined plane slowed the event, while repetition and weighing made a temporal comparison possible without a modern stopwatch. The published report is evidence of Galileo Galilei's experimental presentation, not a surviving raw-data archive from which every reported precision can now be reconstructed.^[ref]

A necessary physical qualification is often missed. A ball in rolling without slipping is not a point particle sliding without friction. For a solid sphere on an ideal inclined plane, modern rigid body dynamics gives $a=5g\sin \alpha /7$, rather than $g\sin \alpha$. Both are constant accelerations, so both give $s\propto t^2$. The trial can therefore test the temporal form without directly establishing the free-fall coefficient. This distinction makes the experiment more intelligible: the evidence supports a specified relation under approximations, rather than every detail of a modern idealization.

The Fourth Day extends this approach to projectile motion. Combine uniform horizontal velocity $u$ with constant downward acceleration $g$, taking downward displacement positive. Then $x=ut$ and $y=g{t}^2/2$. Eliminating time givesa parabola. Actual projectiles experience air resistance, but the ideal curve organizes the effects that need separate treatment. Galileo Galilei thus gave a powerful example of how mathematical abstraction and physical observation could cooperate.^[ref]

Isaac Newton went further by connecting kinds of motion through a general account of force. Newton's second law links the change of momentum to the impressed force; for constant mass, its modern differential expression is $\mathbf{F}=m\mathbf{a}$. Newton's third law makes interaction reciprocal, enabling a consistent analysis of interacting bodies. The Philosophiae Naturalis Principia Mathematica by Isaac Newton develops consequences of central forces and then asks which laws fit astronomical phenomena. Mathematical deduction now connects measurable orbits to an underlying force law, rather than just describing each trajectory separately.^[ref]

A simple modern reconstruction shows the increase in explanatory reach. For a circular orbit of radius $r$ and orbital period $T$ about a much more massive central body, centripetal acceleration is $4{\pi }^{2}r/T^2$. Kepler's third law, $T^2\propto r^3$, therefore implies an acceleration proportional to $r^{-2}$. This circular argument is illustrative; Isaac Newton's treatment of elliptical orbits is more general. With Newton's law of universal gravitation,the two-body orbital period is tied to the semimajor axis $a$ and both masses. The same theory connects the Moon's orbital acceleration with terrestrial gravity, and supports further investigations of tides, comets and departures from a simple two-body orbit.^[ref]

This achievement depended on more than two individuals. Johannes Kepler's relations supplied crucial astronomical constraints, and both authors inherited geometry, observational records and technical practices. Nor should the Philosophiae Naturalis Principia Mathematica by Isaac Newton be described simply as the first application of symbolic calculus: its published arguments are largely geometrical. What distinguishes Isaac Newton is the demonstrative network uniting terrestrial and celestial problems, together with explicit idealizations whose failures generated further calculable questions.

On breadth, quantitative unification and the inference from motion to force, Isaac Newton's contribution was greater. Galileo Galilei's work supplied an exemplary way to mathematize terrestrial motion and test its idealized relations. Giving the former the overall verdict should therefore acknowledge the latter's distinct foundational achievement, not portray the Philosophiae Naturalis Principia Mathematica by Isaac Newton as a creation without predecessors.

For maximum projectile range from an elevated launch, let $X$ be horizontal projectile range from the tower and $\alpha$ the launch angle above horizontal. The projectile motion trajectory isPut $z=\tan \alpha$. For a fixed $X$, its largest possible $y$ occurs at $z=u^2/(gX)$; the envelope is $y_{\max }=h+u^2/(2g)-g{X}^2/(2u^2)$. The furthest attainable ground point satisfies $y_{\max }=0$, soThe optimum angle is less than $45^{\circ }$ for $h>0$ and approaches $45^{\circ }$ as $h\to 0$. The result is the maximum horizontal distance, using uniform gravity and no drag.

Use coordinates $s$ down the inclined plane and $n$ along its upward normal, with the launch point as origin. Gravity has components $g\sin \varphi$ and $-g\cos \varphi$. For launch angle $\theta$ above the downhill direction,The nonzero root of $n(t)=0$ is $t_f=2v_0\sin \theta /(g\cos \varphi )$. Substituting into $s(t)$ and combining the two trigonometric terms gives the distance along the slope:This assumes $0\le \varphi <\pi /2$, $\sin \theta >0$, and a downhill landing with $\cos (\theta -\varphi )>0$, as well as the neglected air resistance. It reduces to the horizontal projectile range $v_0^2\sin (2\theta )/g$ at $\varphi =0$. The angle relative to the horizontal is $\theta -\varphi$, which explains the shifted cosine. The original diagram uses stipulated illustrative angles; no examination figure is being reconstructed.

Take the launch point as $(x,y)=(0,0)$ and ignore air drag. In projectile motion with constant gravitational acceleration $g$, a launch at angle $\theta$ above the horizontal givesAt the wall $x=l$, the height isThis is a concave quadratic polynomial in $q$, so its unique maximum satisfies $l-g{l}^{2}q/v_0^2=0$. HenceFor $l=v_0^2/g$, the best trajectory has $\theta =45^{\circ }$ and reaches the wall at launch height. For $l>v_0^2/g$, the mathematical maximum height is negative. If the ball must stay above level ground, it lands before reaching the wall: $v_0^2/g$ is the maximum same-level projectile range. If there is no intervening ground and the wall extends below the launch height, the same optimum still applies, but its impact point is below that height.

Take horizontal motion towards the second projectile as positive. In projectile motion, both bodies have the same downward gravitational acceleration, so it cancels from their relative motion. Their coordinates areFor a collision at $t>0$, equality of the horizontal coordinates requires $v_2\cos \theta =v_1$. Equality of the heights then requires $v_{2}t\sin \theta =h$. HenceThe upward solution requires $v_2>v_1$. A collision strictly above the ground further requires $t<\sqrt{2h/g}$, or $v_2^2-v_1^2>gh/2$. The horizontal target distance does not independently affect the interception geometry. If the first shot is also required to hit that ground target, its parameters satisfy $l=v_1\sqrt{2h/g}$.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/5/Physics%20Papers%202016.pdf?cookie_passthrough=1

The held cover dates CP1 to 5 June 2017, despite the source archive's filename. All questions and both projectile motion cases are covered. Lettered or Roman divisions are inferred where separate tasks have no printed labels.

Choose upward velocity positive during ascent. Gravity and linear drag both point downward, giving the linear differential equationAt the highest point $v=0$, soThe vertical ascent under linear drag time is smaller than $v_0/g$ for $\alpha >0$, because $\log (1+x)<x$. Its zero-drag limit is $v_0/g$, as required by constant-gravity projectile motion.

For a launch angle $\theta$, the projectile motion equations give a flight time $2u\sin \theta /g$, a projectile range $R=u^2\sin (2\theta )/g$, and an apex height $u^2{\sin }^2\theta /(2g)$. Thus the ceiling requires ${\sin }^2\theta \le 2gH/u^2$.

Set $z={\sin }^2\theta$. Then $R^2=4u^{4}z(1-z)/g^2$, which increases up to $z=1/2$. The optimum is therefore $z=\min (1/2,2gH/u^2)$, giving the projectile range under a ceilingThe printed expression is valid in the first regime. A higher ceiling leaves the usual $45^{\circ }$ launch admissible, so it cannot reduce the maximum range. The angle in the ceiling-limited regime is $\arcsin \sqrt{2gH/u^2}$. Equality at the roof is allowed in this ideal geometrical model; if touching the roof is forbidden, the same value is a supremum approached from below.

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertiawhere $r_{\perp }$ is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density $\lambda =m/l$,The second result uses the parallel axis theorem; direct integration over $0\le x\le l$ gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let $w$ be the particle's signed outgoing horizontal velocity, and $\omega$ the rod's outgoing angular velocity, positive towards the incident motion. ThenSubstituting $\omega =3(v_0-w)/l$ into the kinetic energy equation givesDiscarding the zero-impulse branch $w=v_0$, the actual collision givesThe rod's lower end initially moves forward at $l\omega =3v_0/2$, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and $z$ downward, its trajectory is $x=(v_0/2)t$, $z=l+g{t}^2/2$, where $t=0$ at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is $I_{\mathrm{end}}{\omega }^2/2=3m{v}_0^2/8$. Reaching the horizontal raises its centre of mass by $l/2$, increasing gravitational potential energy by $mgl/2$. Consequently,For the threshold case, measure the rod's angle $\theta$ from the downward vertical and reset $t=0$ at its first horizontal turning point, so $\theta (0)=\pi /2$ and $\dot{\theta }(0)=0$. The restoring torque about the pivot gives the exact equation of motionAt the horizontal, $\ddot{\theta }=-3g/(2l)$, so the rod immediately swings back. Conservation of mechanical energy yieldsOn the initial descent its explicit time dependence is specified by the definite integralIt continues through the vertical to $-\pi /2$, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put $k=1/\sqrt{2}$, ${\omega }_r=\sqrt{3g/(2l)}$, and let $K(k)$ denote the complete elliptic integral of the first kind. ThenThis is not simple harmonic motion at an amplitude of $\pi /2$. Its oscillation period exceeds its own small-amplitude value $2\pi \sqrt{2l/(3g)}$ by the factor $2K(1/\sqrt{2})/\pi \simeq 1.18034$.

A simple pendulum of length $l/2$ obeys $\ddot{\theta }+(2g/l)\sin \theta =0$. At the same amplitude $\pi /2$, its oscillation period is $T_p=4\sqrt{l/(2g)}K(1/\sqrt{2})$, soThe rod swings more slowly. Its equivalent simple pendulum length is $I_{\mathrm{end}}/[m(l/2)]=2l/3$, not the distance $l/2$ to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

With zero gravitational potential energy at the wheel axis, the release energy and the energy at the apex of the projectile motion areOnly the vertical velocity vanishes at the apex: the horizontal kinetic energy remains. Equating the two energies gives

Powered vertical flight. Take upward velocity $v$ as positive, and let $v_e>0$ be the downward exhaust speed relative to the rocket. During $dt$ the rocket's mass changes by $dm<0$, and exhaust mass $-dm$ leaves with Earth-frame velocity $v-v_e$. Compare the momentum of the rocket plus this newly expelled exhaust before and after the interval:Discard the second-order product $dmdv$. The result is $mdv+v_{e}dm=-mgdt$, orThe positive rocket thrust is $-v_e\dot{m}$. Integrating with $v(0)=0$ and $m(0)=m_0$ gives the rocket equation with gravitational loss:Using $d(mv)/dt$ for the rocket alone without the exhaust momentum flux would give the wrong equation for this open system.

Orbital speed under the stated gravity assumption. A circular orbit of radius $r_o=R_E+h_o=7.90\times 10^6\mathrm{m}$ requires centripetal acceleration $v_o^2/r_o$. If the magnitude of gravitational acceleration is the stipulated constant $g=9.8\mathrm{m}{\mathrm{s}}^{-2}$ even at the orbit,This treats gravity as centrally directed for the orbital calculation, while retaining its constant magnitude. The subsequent flat-surface approximation treats its direction as locally vertical.

Burnout height. During the unpowered projectile motion, horizontal velocity is constant. At the top of the trajectory the vertical component vanishes, so the horizontal component must equal the orbital speed $v_o$. At burnout, therefore,These follow either by integrating $\ddot{y}=-g$ or by the vertical work-energy theorem. Substituting the preceding constant-$g$ orbital result givesThe vertical component is initially positive, as required to reach the orbit after burnout.

Alternative inverse-square interpretation. A real Earth's exterior Newtonian gravitational field is not constant over a $1500\mathrm{km}$ altitude change. If $9.8\mathrm{m}{\mathrm{s}}^{-2}$ instead means the surface value $g_0$, then Newton's law of universal gravitation gives $GM=g_0{R}_E^2$ andUsing this orbital speed but the paper's constant-$g$ flat projectile motion approximation for the coast would giveThis alternative changes the global constant-gravity assumption; it is not the same model as the $183\mathrm{km}$ result. An exact curved Kepler orbit coast would also abandon the prescribed flat trajectory. The printed assumptions support the constant-magnitude answers above, while the inverse-square values explain the different answer obtained from the usual physical Earth-orbit formula.

An object launched into motion whose subsequent trajectory is determined by external forces. The ideal elementary projectile motion model retains uniform gravity and neglects air resistance.

The horizontal displacement between launch and landing in a projectile motion model. With equal launch and landing heights, uniform gravitational acceleration $g$, no air resistance, launch speed $u$ and angle $\theta$, it is $R=u^2\sin (2\theta )/g$. Different landing heights or forces change the result.