Kinematics

Take the point of projection as the origin, with $x$ horizontal and $y$ vertically upward. The projectile is launched with speed $V$ at angle $\theta$.

Initial components: $V_x=V\cos\theta$, $V_y=V\sin\theta$.

Horizontal motion (no acceleration): $x=V\cos\theta\,t\;\Rightarrow\; t=\dfrac{x}{V\cos\theta}$

Vertical motion (acceleration $-g$): $y=V\sin\theta\,t-\tfrac12gt^2$

Substituting $t$:

$y=V\sin\theta\cdot\frac{x}{V\cos\theta}-\frac{g}{2}\frac{x^2}{V^2\cos^2\theta}$

$\boxed{y=x\tan\theta-\frac{g\,x^2}{2V^2\cos^2\theta}}$

This has the form $y=ax-bx^2$, which is a parabola, so the path of a projectile is a parabola.

Velocity components at any point (time $t$):

• horizontal: $v_x=V\cos\theta$ (constant)
• vertical: $v_y=V\sin\theta-gt$
• resultant: $v=\sqrt{v_x^2+v_y^2}$, directed along the tangent to the path at angle $\beta$ with $\tan\beta=v_y/v_x$.

Diagram: draw a parabola rising from the origin at angle $\theta$ to the $x$-axis, with the initial velocity $V$ drawn along the tangent at the origin and resolved into $V\cos\theta$ (along $x$) and $V\sin\theta$ (along $y$). At a general point P on the curve draw $v$ along the tangent, with its components $v_x=V\cos\theta$ horizontal and $v_y$ vertical (upward on the rising part, zero at the highest point, downward on the falling part).

When the body is released it already has the horizontal velocity of the aeroplane. After release it moves as a projectile: it keeps moving forward horizontally at that speed while falling under gravity, so it follows a parabolic path and lands ahead of the point of release. If it were dropped exactly above the target, it would overshoot. So it must be released early, when the plane is still a horizontal distance $x=v\sqrt{2h/g}$ short of the target, to hit it.

(b) horizontal component of velocity

With air resistance neglected, the only force is the vertical weight, so there is no horizontal acceleration and $v_x=u\cos\theta$ stays constant. The vertical component, kinetic energy and potential energy all change along the path.

The ball returns to its release height after one range $R$ and one time of flight $T$.

$R=\frac{u^2\sin2\theta}{g}=\frac{20^2\times\sin90^\circ}{9.8}=40.8\ \text{m}$

$T=\frac{2u\sin\theta}{g}=\frac{2\times20\times\sin45^\circ}{9.8}=2.89\ \text{s}$

The player stands 50 m from the thrower, but the ball lands only 40.8 m away. The player must therefore cover

$d = 50-40.8 = 9.2\ \text{m}$

towards the thrower, in time $T$:

$v=\frac{d}{T}=\frac{9.2}{2.89}\approx 3.2\ \text{m/s}$

So the player must run at about $3.2$ m/s along the line towards the thrower (back towards where the ball was thrown from).

(a) From $v=u+at$ with $u=0$:

$t=\frac{v}{a}=\frac{75}{2.5}=30\ \text{s}$

(b)

$s=\frac{v^2}{2a}=\frac{75^2}{2\times2.5}=1125\ \text{m}$

(c) The runway left is $2200-1125=1075$ m.

During the 2.5 s reaction time the aircraft keeps going at 75 m/s:

$s_1=75\times2.5=187.5\ \text{m}$

Then braking from 75 m/s to rest at $4\ \text{m s}^{-2}$:

$s_2=\frac{v^2}{2a}=\frac{75^2}{2\times4}=703\ \text{m}$

Total stopping distance $=187.5+703=891$ m, which is less than 1075 m. Yes, the aircraft can be stopped on the runway, with about 184 m to spare.

Time of flight: the total time for which a projectile remains in the air, from the instant of projection until it returns to the level of projection. For projection speed $u$ at angle $\theta$, $T=\dfrac{2u\sin\theta}{g}$.

Horizontal range: the horizontal distance between the point of projection and the point where the projectile returns to the same horizontal level. $R=\dfrac{u^2\sin 2\theta}{g}$.

(i) Vertically the ball starts with zero velocity, so it falls freely through $h=24$ m:

$h=\tfrac12 g t^2\;\Rightarrow\; t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times24}{9.8}}\approx 2.2\ \text{s}$

(ii) The horizontal velocity stays constant (no air resistance), so $x=ut$:

$u=\frac{x}{t}=\frac{27}{2.21}\approx 12\ \text{m s}^{-1}$

The ball was thrown at about $12\ \text{m s}^{-1}$.

(b) The athlete has a constant horizontal velocity and constant downward acceleration.

In projectile motion the only force acting (air resistance being negligible) is weight, which acts vertically downward. So the horizontal velocity stays constant throughout the jump, while the vertical motion has a constant downward acceleration equal to $g$.

(a) From the velocity–time graph, the box speeds up uniformly from $0$ to $5\ \text{m s}^{-1}$ in the first $2.0\ \text{s}$.

$a=\frac{v-u}{t}=\frac{5-0}{2.0}=2.5\ \text{m s}^{-2}$

With the unbalanced force $F=10\ \text{N}$, Newton's second law gives

$m=\frac{F}{a}=\frac{10}{2.5}=4\ \text{kg}$

(b) After $2.0\ \text{s}$ the graph is horizontal, so the velocity is constant and $a=0$. The net force is therefore zero, and friction balances the $12\ \text{N}$ pull:

$f=12\ \text{N}$

(c) Distance in $8.0\ \text{s}$ = area under the graph from $0$ to $8.0\ \text{s}$:

$s=\tfrac12\times 2.0\times 5+(8.0-2.0)\times 5=5+30=35\ \text{m}$

To reach the point directly opposite, the boat must have no net component of velocity along the river. The boatman should therefore row upstream, tilted at an angle $\theta$ to the line perpendicular to the bank.

Let $v_b$ be the boat's speed in still water and $v_r$ the speed of the river. The component of $v_b$ along the river (upstream) must cancel the river current:

$v_b\sin\theta=v_r\;\Rightarrow\;\theta=\sin^{-1}\!\left(\frac{v_r}{v_b}\right)$

Diagram: draw the river flowing to the right, the bank-to-bank line AB perpendicular to the banks, the river velocity $\vec v_r$ along the river, the boat's heading $\vec v_b$ pointing upstream at angle $\theta$ from AB, and the resultant velocity along AB. This works only if $v_b>v_r$, and the crossing speed is $v_b\cos\theta$.