Shot put

Equation in horizontal direction: $23=x=v*cos(\theta)\times t$ let $t$ equal time it takes to travel $23m$

Equation in vertical direction: $0=y=-4.9t^2+v*sin(\theta)\times t+1.67$ . It is zero because at $t$ , ball is back on the ground.

Kinetic energy is $\frac{1}{2} \times m*v^2$ . Since mass is given, it is obvious that $v^2$ must be minimized.

We know the trig identity $v^2=(v*cos(\theta))^2+(v*sin(\theta))^2$ . Now, the trick is that both kinematic equations can be solved for $v*cos(\theta)$ and $v*sin(\theta)$ respectively, and without much difficulty.

$vcos(\theta)=\dfrac{23}{t}$

$vsin(\theta)=\dfrac{4.9t^2-1.67}{t}$

Substituting the kinematic equation into the trig identity, we get $\dfrac{(4.9t^2-1.67)^2+23^2}{t^2}$ . Using Wolfram Alpha or using the Quotient Rule and setting the derivative to zero, we get the minimum square of the velocity to be $\dfrac{49}{500} \times (17\sqrt(18401)-167)$ , or roughly $209.627$

K.E= $\frac{1}{2} \times 7.26 kg \times 209.627 \frac{m^2}{s^2} =760.947\approx760.950$

Note that, if $v$ is the original velocity and $\theta$ is the initial projectile angle, we have (ignoring units): $h = \dfrac {(v \sin \theta)^2}{2g}$ and $\left[ \sqrt {\dfrac {2 \cdot (h+1.67)}{g}} + \dfrac {v \sin \theta}{g} \right] \cdot \left( v \cos \theta \right) = 23.$ Substituting, we get $\left[ \sqrt {\dfrac {1}{4.9} \cdot \left( 1.67 + \dfrac {v^2 \sin^2 \theta}{19.6} \right)} + \dfrac {v \sin \theta}{9.8} \right] \cdot \left( v \cos \theta \right) = 23.$ Solving for $v$ , we get $v \approx \dfrac {5.92663 \cdot 10^7 \cdot \sec \left( \theta \right)}{\sqrt {3.11668 \cdot 10^{13} \cdot \tan \left( \theta \right) + 2.26298 \cdot 10^{12}}}$ We find, from this, that $\text {min} (v) = 14.48 \, \text{m/s} \rightarrow \text {min} (KE) = \boxed {761 \, \text{J}}$ .

We seek to minimize the initial kinetic energy, $\frac{1}{2}mv_0^2$ Since $m$ is constant, we factor that out and attempt to find the minimum value of $\frac{1}{2}v_0^2$ .

Let $\theta$ be the launch angle of the shot put. We have two equations from essential kinematics formulas:

$-4.9t^2 +( v_0 \sin \theta) \, t = -1.67$ (1)

$v_0 \cos \theta = \frac{23}{t}$ (2)

Rearranging (2) we have:

$t = \frac{23}{v_0 \cos \theta}$

Substituting into (1):

$\frac{(-4.9)(23)^2}{v_0^2 \cos^2 \theta} + \frac{23 v_0 \sin \theta}{v_0\cos \theta} = -1.67$

Rearranging and solving for $v_0^2$ :

$v_0^2 = \frac {2592.1}{(\cos^2 \theta)(23 \tan \theta + 1.67)}$

Dividing by two:

$\frac{1}{2} v_0^2 = \frac {1296.05}{(\cos^2 \theta)(23 \tan \theta + 1.67)}$

Now use calculus and a calculator or a CAS to find the minimum of the above. Multiply by $m$ , $7.26$ at the end!

$For\quad motion\quad in\quad x\quad direction,\quad we\quad have\quad the\quad following\quad equation:\\ \\ \\ \\ \qquad \qquad v\quad cos\theta \quad t\quad =\quad 23\\ \\ \\ And\quad for\quad motion\quad in\quad y\quad direction,\quad we\quad have\quad got\quad this\quad equation:\\ \\ \\ \qquad \qquad -1.67\quad =\quad v\quad sin\theta \quad t\quad -\quad \frac { 1 }{ 2 } 9.8\quad { t }^{ 2 }\\ \\ \\ The\quad problem\quad is\quad asking\quad to\quad minimize\quad the\quad value\quad of\quad v.\\ I\quad did\quad this\quad usind\quad online\quad calculator\quad but\quad can\quad be\quad found\quad by\quad hand\quad too\\ but\quad it\quad would\quad be\quad more\quad like\quad a\quad maths\quad question.\\ Physics\quad ends\quad right\quad here.$

Well first off, we know that we will need to give the shot put the least amount of kinetic energy when the angle is exactly at 45 $^{o}$ (At least most people have come across that problem and found that solution). With that knowledge, we can derive the equations needed to solve for the missing variables.

Let the upwards direction be +y and the direction parallel with the floor and to the right be +x . (I'm assuming the shot put was thrown to right for no particular reason). Let the initial velocity of the ball at angle $\theta$ from the ground be $V_{0}$ . The shot put is initially at $x = 0$ and $y = 1.67$

*+y : * $\frac{-gt^{2}}{2} + V_{0} \sin \theta t + 1.67 = 0$

*+x : * $V_{0}\cos\theta t = 23$

Now you've probably noticed that $t$ is the variable that represents the time elapsed, and in this case, it is the time elapsed once the ball has hit the ground. Let's solve for $V_{0}$ .

$t = \frac{23}{V_{0}\cos\theta}$ (derived from the +x formula)

Plugging it into the +y formula...

$\frac{-g(23^{2})}{2(\cos^{2}\theta V_{0}^{2})} + \frac{23(V_{0} \sin \theta)}{V_{0} \cos \theta} + 1.67 = 0$

$\frac{529g}{2\cos^{2}\theta V_{0}^{2}} = 23\tan\theta + 1.67$

$V_{0}^{2} = \frac{529g}{2\cos^{2}\theta(23\tan\theta + 1.67)}$

Plugging in $9.81$ in for $g$ and $45^{o}$ or $\frac{\pi}{4}$ in for $\theta$ , you will get $V_{0}^{2} = 210.3563$ .

The kinetic energy of an object travelling very slowly compared to the speed of light can be modeled by the following equation.

$KE = \frac{1}{2}mv^{2}$

Plugging in $V_{0}^{2}$ for $v$ and $7.26$ for $m$ we get...

$KE = \frac{1}{2}(7.26)(210.3563) = \boxed{760.95}$

v \times \sin 45 \times t - 4.9 \times t^{34} = -1.67m

v \times \cos 45 \times t = 23m

Solving for t you get

** t = \frac {23} {v \times \cos 45}

Then solving for v through substitution you get

* v = 14.49 m/s *

Applying KE = 0.5 m v^{2}

You get:

\boxed {761}