Minimum speed

Classical Mechanics Level 5

A particle projected from the ground passes two points, which are at height $h_{1}=12 m$ and $h_{2}=18 m$ above the ground and at a horizontal distance $d=8 m$ apart (diagonal distance $10 m$ apart). What could be the minimum speed $u$ of projection?

$g =10 m/s^2$ . All other force are neglected.
$u$ is in $m/s$

The answer is 20.00.

3 solutions

Mark Hennings
Sep 20, 2016

If the initial horizontal and vertical components of the particle's velocity at ground level are $U,V > 0$ respectively, then the particle has trajectory $y \; = \; \tfrac{V}{U}x - \tfrac{g}{2U^2}x^2$ Suppose that the particle passes through two points, which are heights $h_1$ and $h_2$ above the ground and a distance $D$ apart. Thus the horizontal distance between these points is $d$ , where $(h_1-h_2)^2 + d^2 = D^2$ . Thus we deduce that $h_1 \; =\; \tfrac{V}{U}X - \tfrac{g}{2U^2}X^2 \hspace{2cm} h_2 \; = \; \tfrac{V}{U}(X+d) - \tfrac{g}{2U^2}(X+d)^2$ for some $X > 0$ . We may assume, without loss of generality, that the particle meets the height $h_1$ point first.

The first of these equations tells us that $X \; = \; \frac{UV \pm U\sqrt{V^2 - 2gh_1}}{g}$ whereas the difference of the two equations tells us that $\begin{array}{rcl} h_2 - h_1 & = & \tfrac{V}{U}d - \tfrac{g}{U^2}dX - \tfrac{g}{2U^2}d^2 \\ X & = & \tfrac{1}{g}UV - \tfrac12d - \frac{(h_2-h_1)U^2}{gd} \end{array}$ Comparing these two results tells us that $\begin{array}{rcl} \tfrac{U}{g}\sqrt{V^2 - 2gh_1} & = & \tfrac12d + \frac{(h_2-h_1)U^2}{gd} \\ V^2 & = & 2gh_1 + \frac{(gd^2 + 2(h_2-h_1)U^2)^2}{4d^2U^2} \; = \; \frac{(h_2-h_1)^2}{d^2}U^2 + g(h_1+h_2) + \frac{g^2d^2}{4U^2} \\ U^2 + V^2 & = & \frac{D^2}{d^2}U^2 + g(h_1+h_2) + \frac{g^2d^2}{4U^2} \; \ge \; g(h_1+h_2) + 2\sqrt{\tfrac14g^2D^2} \; =\; g(h_1+h_2+D) \end{array}$ using the AM/GM inequality. Thus the minimum possible projection speed is $\sqrt{g(h_1+h_2+D)} = \boxed{20}$ in this case.

The question says distance between the towers is 10 m , i think it implies that the horizontal distance between the towers is 10 m .

space sizzlers - 4 years, 7 months ago

The question does not mention towers! It mentions two points, and describes their vertical heights and the distance between them. If there were towers, it might be reasonable to think of the horizontal distance between them. Since there are no towers. the distance between two points is the straight line distance.

Mark Hennings - 4 years, 7 months ago

thank you....understood .

space sizzlers - 4 years, 7 months ago

I also interpreted the phrase "at a distance d=10 m apart" to mean the horizontal separation down-range. The answer, in that case, is a minimum launch velocity of 21 m/s at an angle of 76.88625 degrees from horizontal. But as Mark's solution and comments state, the "distance apart" is to be taken as "straight-line distance between the points". Then the solution for minimum speed is indeed 20 m/s at a launch angle of 73.56267 degrees. I just think making it clear whether the "10 m apart" refers to horizontal or diagonal measurement would make the question better. On the shooting range I visit, they refer to 'distance apart of targets at different heights above the ground', as the "distance along the ground". That threw me off.

Bob Kadylo - 4 years, 5 months ago

Tushar Gautam
Aug 26, 2014

Answer :- $\sqrt {g (d +h_{1} +h_{2})} = 20 m/s$

Let particle projected from ground with speed $u$ and angle $\theta$ with positive x-axistry to and when it reach at height $h_{1}$ let its speed be $v$ and make angle $\alpha$ with positive x-axis.

Now $v_{y}^2 = u_{y}^2 -2 g h_{1}$ because $a_{y} = - g$

And $v_{x}^2 = u_{x}^2$ because $a_{x} = 0$

So adding gives $u^2 = v^2 + 2 g h_{1}$ implies in order to minimize $u$ we need to minimize $v$

After reaching $h_{1}$ particle has to reach height $h_{2}$ while covering horizontal distance of $d$ so we can assume a inclined plane made by joining $h_{1}$ and $h_{2}$ and the particle has to reach from $h_{1}$ to $h_{2}$ with minimum speed $v$ making an angle $\alpha$ with positive x-axis

We may prove it latter that for a speed to maximum the range on an inclined plane making a angle $\theta_{in}$ with positive x axis, the angle made by particle with the inclined plane is given by $\alpha^' = \frac {\pi}{4} - \frac {\theta_{in}}{2}$

In this case $tan \theta_{in} = \frac {h_{2}-h_{1}}{d}$ ,

time of flight from $h_{1}$ to $h_{2}$

$t = \frac {2 v sin(\alpha^')}{g cos(\theta_{in})}$

$R = \sqrt {d^2 + (h_{2}-h_{1})^2} = v cos (\alpha^') t - \frac {g sin (\theta_{in})}{2} t^2$ $= \frac {v^2}{g cos^2(\theta_{in})} \times [sin (\theta_{in} + 2 \alpha^') -sin (\theta_{in})]$ one step simplfied.

From here we can prove if $\theta_{in} + 2 \alpha^' = \frac {\pi}{2}$ range is maximum ,this gives the above result for angle.

$v^2 = \sqrt {d^2 + (h_{2}-h_{1})^2} \frac {g cos^2(\theta_{in})}{1- sin \theta_{in}}$

I 'll continue latter maybe I have done mistake their and how to upload a photo.

I have a different solution:

$u\approx{19.05m/s}$

$h_1=18m,\,h_2=18m,\,g=10m/s^2$

$d=10m,\,x_1=x,\,x_2=x+d$

${\dot x}={\dot {x_0}}=u\cos{\theta},\,x(0)=0,\,x=ut\cos{\theta},\,(1)$

${\ddot y}=-g,\,{\dot {y_0}}=u\sin{\theta},\,y(0)=0,\,{\dot y}=- gt+u\sin{\theta}$

$y=-{\frac 1 2}gt^2+ut\sin{\theta},\,(2)$

substituting $t={\frac x {u\cos\theta}}\,$ in the equation $(2)$ , it can obtained the equation of the trajectory:

$y=-{\frac 1 2}g{\frac {x^2} {u^2{{\cos}^2}\theta}}+x\tan\theta,\,(3)$

The law of conservation of energy implies:

${\frac 1 2}mv^2+mgy={\frac 1 2}mu^2$

$v^2={\dot {x_0}}^2+{\dot y}^2=u^2{\cos}^2\theta+{\dot y}^2$

simplyfing:

$u^2{\cos}^2\theta+2gh_2=u^2,\,u^2{\sin}^2\theta=2gh_2,\,u\sin\theta=\sqrt{2gh_2}\,(4)$

because the traiectory is a parabola, the minimum speed $u$ is reached when the vertex is in $x_2=x+d,\,$ and $y(x_2)=h_2,\,{\dot y}(x_2)=0$

$-gt+u\sin{\theta}=0,\,t_2=\frac {u\sin\theta} g=\frac {x+d} {u\cos\theta}$

$x+d={\frac {u^2\sin\theta\cos\theta} g}$

$x={\frac {{u^2}{\sin\theta}{\cos\theta}} g}-d$

$\sin\theta={\frac{\sqrt{2gh_2}} u},\,\cos\theta=\sqrt{1-{\sin}^2\theta}={\frac {\sqrt{u^2-2gh_2}} u}$

$\tan\theta={\frac {\sin\theta}{\cos\theta}}={\sqrt{\frac {2gh_2}{u^2-2gh_2}}}$

we must impose the condition:

$u^2>2gh_2,\,u>u^{*}=\sqrt{2gh_2}=\sqrt{360}\approx {18.97}$

imposing the condition $y(x)=h_1,\,$ we get:

$h_1=-{\frac 1 2}g{\frac {x^2} {u^2{{\cos}^2}\theta}}+x\tan\theta$

substituting the values for $h_1,\,x,\,\sin\theta,\,\cos\theta,\,\tan\theta\,:$ and set $u^2-2gh_2=u^2-360=z,\,$ we obtain:

$h_1={\frac g {2z^2}}{({\sqrt{\frac {2h_2} g}}z-d)^2}$

$12={\frac 5 {z^2}}{({\sqrt{\frac {18} 5}}z-10)^2}$

$({z-5{\sqrt {10}}})^2={\frac {500} 3}$

$z_{1,2}=5\sqrt 10 \pm 10\sqrt \frac 5 3$

or by wolfram alpha we get:

$u^2-360=z_{1,2}$

$u_1\approx {19.05},\,u_2\approx {19.72}$

so the minimum value is:

$u\approx \boxed {19.05}$

Antonio Fanari - 6 years, 8 months ago

@Mark Hennings , sir this solution has mistakes can you please correct it?

space sizzlers - 4 years, 8 months ago

I may have a much easier solution to this i.e 2 to 3 lines solution. I will post it as soon as I can.....

Tushar Gopalka - 6 years, 8 months ago

Manas Verma
Aug 21, 2014

use the formula time of descent=sq.2h/g max.height of ball=18+(12-10)=20 time of decent is known so use t=u/gand calculate the value og u. ANS. U=20m/s

How you got maximum height of ball as $18+(12-10)$ or $h_{2}+h_{1}-d$

tushar gautam - 6 years, 9 months ago

Hey tushar please provide a solution to your question

space sizzlers - 6 years, 9 months ago

Please explain how maximum height of ball is 20

space sizzlers - 6 years, 9 months ago

Nice question but solution needed tushar

swapnil rajawat - 6 years, 9 months ago

Minimum speed

The answer is 20.00.

3 solutions

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