July 25, 2019 – Physicspupil

The kinetic energy $E_\mathrm{k}$ of an object of mass $m$ and speed $v$ is given by the formula

$E_\mathrm{k}=\displaystyle{\frac{1}{2}}mv^2$ .

In the situation that the object is thrown upwards with initial speed $u$ , and subjected only to gravity $\mathbf{g}$ , it can be expected that after some time of flight $T$ , the object will return to its initial position, its downward speed in which is equal to the initial upward speed $u$ .

Define a piecewise scalar function $v(t)$ of time $t$ :

$v(t) = \begin{cases} u-gt & \quad \textrm{when }0\leq t\leq \displaystyle{\frac{T}{2}} \\ -u+gt & \quad \textrm{when } \displaystyle{\frac{T}{2}}\leq t\leq T \end{cases}$

or simply

$v:[0,T]\subset \mathbb{R} \rightarrow [0,u]\subset \mathbb{R}$ given by $t\mapsto \big|u-g(T-t)\big|$ .

Then

$\begin{aligned} v^2 & =(u-gt)^2\quad \big( =(-u+gt)^2\big) \\ & = u^2-2ugt+g^2t^2 \quad \big(\forall\, t\in [0,T] \big) \end{aligned}$ .

Thus the kinetic energy $E_\mathrm{k}(t)$ is

$\begin{aligned} E_\mathrm{k}(t) & =\displaystyle{\frac{1}{2}}m(u^2-2ugt+g^2t^2) \\ & = \bigg( \displaystyle{\frac{1}{2}}mg^2 \bigg) t^2 + ( -mug ) t + \bigg( \displaystyle{\frac{1}{2}}mu^2 \bigg) \end{aligned}$

where $m$ , $u$ , $g$ are constants.

The kinetic energy $E_\mathrm{k}(t)$ is set to zero at time $t'$ :

$\begin{aligned} t' & = \displaystyle{\frac{-(-mug)\pm\sqrt{\big( -mug\big)^2-4\big(\frac{1}{2}mg^2\big)\big(\frac{1}{2}mu^2\big)}}{2\big(\frac{1}{2}mg^2\big)}} \\ & = \displaystyle{\frac{mug\pm\sqrt{m^2u^2g^2-m^2u^2g^2}}{mg^2}} \\ & =\displaystyle{\frac{u}{g}} \end{aligned}$

Substituting $t'=\displaystyle{\frac{u}{g}}$ for $t$ in $v=u-gt$ :

$v=u-g\bigg( \displaystyle{\frac{u}{g}} \bigg) =0$ ,

as checked.

And the answer is B.

When projected in the air and subjected only to gravity, a projectile performs projectile motion. Its path/trajectory is a parabola. If the projectile is projected at an angle of projection, it is said to be in general projectile motion; else, it is said to be in horizontal projectile motion.

The duration of time from projection to landing is called the time of flight. The maximum height the projectile can reach is usually measured from the launch level. The range of projection is the horizontal distance the projectile has travelled from projection to landing.

Horizontally Projected Motion

When air resistance is negligible, a projectile moves at a uniform horizontal velocity and at a uniform vertical acceleration due to gravity. As its motion in the horizontal and in the vertical are independent of each other, so we separate the projectile motion into two perpendicular directions—the horizontal and the vertical—in order to resolve and analyse it.

i. ; ii. ; iii. ; iv. are what you need.

At any instant $t$ :

horizontal motion ( $u_x=u$ , $a_x=0$ ):

$v_x=u_x=u$ ;

$s_x=u_xt=ut$ OR $t=\displaystyle{\frac{s_x}{u}}$

vertical motion ( $u_y=0$ , $a_y=-g$ ):

$v_y=u_y+a_yt=0-gt=-gt$ ;

$s_y=u_yt+\frac{1}{2}a_yt^2=0+\frac{1}{2}(-g)t^2=-\frac{1}{2}gt^2$ ;

$v_y^2=u_y^2+2a_ys_y=0+2(-g)s_y=-2gs_y$ ;

the velocity $v$ can be found by

$\sqrt{v_x^2+v_y^2}=\sqrt{u^2-2gs_y}\enspace (=\sqrt{u^2+g^2t^2})$

its trajectory is parabolic: $s_y=\displaystyle{-\frac{g}{2u^2}s_x^2}$ .

General Projectile Motion

A projectile is now projected with an initial velocity $u$ at an angle $\theta$ . As usual, we separate the motion into horizontal and vertical directions. At any instant $t$ , the magnitude of its velocity is $v=\sqrt{v_x^2+v_y^2}$ , and its direction makes an angle $\phi$ with the level, where $\tan\phi =\displaystyle{\frac{v_y}{v_x}}$ .

Horizontal motion: The initial horizontal velocity is $u_x=u\cos\theta$ . After some time $t$ , its horizontal displacement is $s_x=u_xt+\frac{1}{2}a_xt^2$ OR $s_x=u\cos\theta t$ and its velocity remains to be $v_x=u_x+a_xt$ OR $v_x=u_x=u\cos\theta$ .

(Neglecting air friction, we assume zero horizontal acceleration, i.e., $a_x=0$ .)

Vertical motion: The initial vertical velocity is $u_y=u\sin\theta$ . After some time $t$ , its vertical displacement is $s_y=u_yt+\frac{1}{2}a_yt^2$ OR $s_y=u\sin\theta t-\frac{1}{2}gt^2$ and its velocity changes to $v_y=u_y+a_yt$ OR $v_y=u\sin\theta -gt$ .

(Upward taken to be +ve, the vertical acceleration due to gravity is $a_y=-g$ )

Better to know the equations of motion well in deriving, than to simply memorize the ~~formulae~~ in solving, the unknowns, lest it be wrong in some cases, say, on a slant.

Time of flight $t$ : Consider only the vertical motion, by ①: $s_y=u_yt+\displaystyle{\frac{1}{2}}a_yt^2$

$\begin{aligned} 0&=(u\sin\theta)t-\frac{1}{2}gt^2 \\ 0&=t(u\sin\theta -\frac{1}{2}gt) \\ t&= 0\quad \mathrm{or}\quad \boxed{t=\frac{2u\sin\theta}{g}} \end{aligned}$

Or, by ②: $v=u+at$ , and that the object lands with the same speed as is launched, we have $-u\sin\theta =u\sin\theta -gt$ , which also gives $t=2u\sin\theta /g$ .

Range $R$ : Consider only the horizontal motion, by $s_x=u_xt$ ( $\because a_x=0$ )

$\begin{aligned} R&=(u\cos\theta )\bigg( \frac{2u\sin\theta}{g}\bigg) \\ &=\frac{2u^2\sin\theta\cos\theta}{g} \end{aligned}$

$\boxed{R=\frac{u^2\sin 2\theta}{g}}$

is maximum if (). . Both angles and give the same range.

Maximum height $H$ : Consider only the vertical motion, by ①: $v_y^2=u_y^2+2a_ys_y$

$0^2=(u\sin\theta )^2-2gH$

$\boxed{H=\frac{u^2\sin^2\theta}{2g}}$

Or, by ②: $s_y=u_yt+\frac{1}{2}a_yt^2$ and that it takes half of the time of flight to reach $H$ , we have $H=(u\sin\theta)\bigg( \displaystyle{\frac{u\sin\theta}{g}}\bigg)+\displaystyle{\frac{1}{2}}(-g)\bigg( \displaystyle{\frac{u\sin\theta}{g}}\bigg)^2$ and thus the same result.

CONCEPT TEST

As shown in the figure, three objects , , and are projected horizontally and travelled along their respective trajectories. Objects and are projected at the same height. Neglecting air resistance, which of the following statements are true?
1. The time of flight of $a$ is longer than that of $b$ .
2. The time of flight of $b$ is equal to that of $c$ .
3. The horizontal velocity of $a$ is less than that of $b$ .
4. The initial velocity of $b$ is greater than that of $c$ .
1. (I) and (II)
2. (I) and (IV)
3. (II) and (III)
4. (II) and (IV)
As shown in the figure, two objects and are in general projectile motion. They reach the same maximum height. Neglecting air resistance, which of the following statements are wrong?
1. The acceleration of $B$ is greater than that of $A$ .
2. The time of flight of $B$ is equal to that of $A$ .
3. The velocity of $B$ is equal to that of $A$ when they are at the maximum height.
4. The velocity of $B$ is greater than that of $A$ at the time of landing.
1. (I) and (II)
2. (I) and (III)
3. (II) and (III)
4. (III) and (IV)
An object is now in general projectile motion. Which of the following graphs are correct?
1. (I) and (II)
2. (I) and (III)
3. (II) and (IV)
4. (III) and (IV)

Answers:

Explanations:

Since $h=\displaystyle{\frac{1}{2}}gt^2$ , time of flight is given by $t=\displaystyle{\sqrt{\frac{2h}{g}}}$ . From $h_b=h_c>h_a$ , we get $t_a<t_b=t_c$ . Thus (I) is wrong and (II) correct. The horizontal velocity is given by $v=\displaystyle{\frac{x}{t}}$ . From $x_a>x_b>x_c$ , we get $v_a>v_b>v_c$ . Thus (III) is wrong and (IV) correct.
The acceleration of $A$ and of $B$ is due to gravity, and is equal to $-g$ (Upward taken to be +ve). Thus (I) is wrong. Let the vertical component of initial velocity be $u_y$ . From $0=u_y^2-2gh$ , we know that $A$ and $B$ have the same magnitude in their vertical component of initial velocity. Thus they have the same time of flight and (II) is correct. From $v_x=\displaystyle{\frac{x}{t}}$ , where $x_B>x_A$ , we know that the horizontal velocity of $B$ is greater than that of $A$ . The velocity at maximum height is $v=\sqrt{v_x^2+v_y^2}=\sqrt{v_x^2}=v_x$ . Thus $v_B>v_A$ and (III) is wrong. Again by resolving components, the velocity at the time of landing is $\sqrt{v_x^2+v_y^2}$ . From $v_y$ being equal and $v_x$ of $B$ greater than that of $A$ , it follows that (IV) is correct.
By the conservation of mechanical energy, $\Delta \mathrm{KE}+\Delta\mathrm{PE}=0$ , i.e., $\bigg( \displaystyle{\frac{1}{2}}mv^2-\displaystyle{\frac{1}{2}}mu^2\bigg) +(mgh-0)=0$ . Arranging it into the form $y=mx+c$ , we have $\mathrm{KE}(=\frac{1}{2}mv^2)=-\mathrm{PE}(=mgh)+\frac{1}{2}mu^2$ . Thus (I) is correct. (II) is wrong because during the flight, there must be a non-zero horizontal component of velocity and thus $\mathrm{KE}\not\equiv 0$ . The maximum height $H=\displaystyle{\frac{u^2\sin^2\theta}{2g}}$ . We can readily fit it into the form $y=ax^2$ , where $y=H$ , $x=u$ , and $a=\displaystyle{\frac{\sin^2\theta}{2g}}$ . Hence (III) is correct. From $s_y=u_yt+\frac{1}{2}a_yt^2$ , it follows that $H=(u\sin\theta )t-\displaystyle{\frac{gt^2}{2}}$ . We can likewise fit it into a parabola $y=ax^2+bx+c$ where $y=H$ , $x=t$ , $a=-g/2$ , $b=u\sin\theta$ , and $c=0$ . The graph (IV) is far from correct.