It's Blocking The Home Run!

Classical Mechanics Level 5


A Baseball player is ready to hit a Home Run. But, he notices a wall , which is at a distance L \displaystyle L meters from him, standing H \displaystyle H meters high.

This player turns out to be the laziest player in the team, but nonetheless very clever. He devices a situation where he can cross the wall with minimum effort (i.e with minimum initial speed).

Find this minimum speed u m i n \displaystyle u_{min} .

Details and Assumptions:
∙ \bullet L = 10 m \displaystyle L = 10m
∙ \bullet H = 5 m \displaystyle H = 5m
∙ \bullet Take g = 9.8 m / s 2 \displaystyle g = 9.8m/s^2


The answer is 12.5923.

Anish Puthuraya by Anish Puthuraya , 24, India

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6 solutions

Discussions for this problem are now closed

Mayank Khetan
Mar 3, 2014

I have 2 solutions to the problem. The first is of course taking the min. speed to be u and angle a, imposing the restriction and then differentiating and substituting which is quite boring and time-consuming. The second and much interesting solution is to imagine an inclined plane from the player to the top of wall. The inclination of the inclined plane is tan-inverse 1/2( Let it be a). For a fixed u, range along the inclined plane is max for angle x=(90-a)/2 (angle from inclined plane). We find the range in terms u, a and x and then substitute range to be root of 125(from pythagoras theorem). Since a and x are already known, we can find u which is approx. 12.59.

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Christopher Boo
Apr 25, 2014

Let u u be the initial speed and θ \theta be the angle respect to the horizontal.

From the problem statement, we have two equations, which is

10 = ( u cos ⁡ θ ) t 10=\big (u\cos{\theta}\big )t

5 = ( u sin ⁡ θ ) t − 1 2 g t 2 5=\big (u\sin\theta\big )t - \frac{1}{2}gt^2

From the first equation, we have that

t = 10 u cos ⁡ θ \displaystyle t=\frac{10}{u\cos\theta}

Substitute into the second equation we have

5 = u sin ⁡ θ ( 10 u cos ⁡ θ ) − 1 2 g ( 10 u cos ⁡ θ ) 2 \displaystyle 5=u\sin\theta\Big (\frac{10}{u\cos\theta}\Big )-\frac{1}{2}g\Big ({\frac{10}{u\cos\theta}}\Big )^2

Express in terms of u u , it is

u = \strut 98 sec ⁡ 2 θ 1 − 2 tan ⁡ θ \displaystyle u=\sqrt{\strut \frac{98\sec^2\theta}{1-2\tan\theta}}

Now, apply calculus then we can find the minimum value of u u , which is 12.5923 \boxed{12.5923} .

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Komal Sai
Feb 28, 2014

first let us think like the lazy player. He is lazy enough, so he will also solve it lazily.Now, differentiating and again substituting is a burden.

Construct a vector diagram with the sides of a triangle as ABC. Apply, pythagorus theorem to triangle ABC and then use AM-GM inequality

AB=ut ; BC=5+1/2gt^2 ; CA=10

If anyone would tell me how to draw, I will provide the VECTOR diagram of triangle ABC ,

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I am getting answer 14m/s. Since the player wants to do the work with minimum effort,the ball will pass just up to the topmost position of the wall.Hence the wall is the max height reached.now H=5m=(V min^2)sin^2/2g.Now at the highest position of the trajectory, the ball covers half of its range.Therefore, {(v min^2)sin2(theta)}/2g=20m=2L.Solving both the equations we get,Theta=45degree.So,Sin(theta)=1/root(2).Hence,V_min comes out 14m/s.What is wrong with it?

Swapnanil Gupta - 7 years, 3 months ago

Consider the effort of the player as a combination of a horizontal velocity u h o r i z o n t a l u_{horizontal} and a vertical velocity u v e r t i c a l u_{vertical} at the start of the trajectory. What is required to be minimized is the total combined effort i.e. ( u h o r i z o n t a l 2 + u v e r t i c a l 2 ) \displaystyle{ \sqrt {({u_{horizontal}}^{2} + {u_{vertical}}^{2})}} . What you are doing is independently minimizing the effort in vertical direction and hoping for the best. Thus you lose your freedom to save effort on the horizontal velocity.

The correct way to do it would be as follows. Imagine that the horizontal velocity is kept fixed ( we shall vary it later) and we wish to vary the vertical velocity for minimum effort. It is clear that by keeping u h o r i z o n t a l u_{horizontal} fixed the ball reaches the x coordinate x = L x = L at a fixed time t = L u h o r i z o n t a l \displaystyle{t = \frac {L}{u_{horizontal}}} . Now it makes sense to reduce u v e r t i c a l u_{vertical} till the ball just passes over the wall i.e. height reached by ball at the wall at time t is H.. Thus we get a relation between u v e r t i c a l u_{vertical} and u h o r i z o n t a l u_{horizontal} .

Now what is to be minimized is the net velocity = u h o r i z o n t a l 2 + u v e r t i c a l 2 \sqrt {u_{horizontal}^{2} + u_{vertical}^{2}} expressed in terms of a single variable.

Reaber John - 7 years, 3 months ago

Thats just how I solved the problem.(L) is half its range.

Swapnanil Gupta - 7 years, 3 months ago

@Swapnanil Gupta – If you solve it correctly you would get the following :

H = u v e r t i c a l t − 1 2 g t 2 \displaystyle{H = u_{vertical} \ t - \frac {1}{2} g t^{2} }

Thus, u v e r t i c a l = H t + 1 2 g t \displaystyle{u_{vertical} = \frac {H}{t} + \frac {1}{2} g t }

Subtituting t = L u h o r i z o n t a l \displaystyle{t = \frac {L}{u_{horizontal}}}

u v e r t i c a l = H u h o r i z o n t a l L + L g 2 u h o r i z o n t a l \displaystyle{ u_{vertical} = \frac {H \ u_{horizontal} }{L} + \frac {Lg}{2\ u_{horizontal}} }

So for any given u h o r i z o n t a l u_{horizontal} , the above relationship should hold for minimum effort.

Now we can still vary u h o r i z o n t a l u_{horizontal} . Therefore we should minimize the net velocity u n e t u_{net} w.r.t. u h o r i z o n t a l u_{horizontal} . where, u n e t = u v e r t i c a l 2 + u h o r i z o n t a l 2 \displaystyle{ u_{net} = \sqrt { u_{vertical}^{2} + u_{horizontal}^{2} }}

We shall minimize u n e t 2 u_{net}^{2} instead of u n e t u_{net} since it makes no difference.

Let f = u n e t 2 \displaystyle {f = u_{net}^{2}} , then f = u h o r i z o n t a l 2 ( 1 + H 2 L 2 ) + L 2 g 2 4 u h o r i z o n t a l 2 + H g \displaystyle { f = u_{horizontal}^{2} \left( 1 + \frac{H^{2}}{L^{2}} \right ) + \frac {L^{2} g^{2}}{4 \ u_{horizontal}^{2}} + H g}

For minimum effort : d f d u h o r i z o n t a l = 0 \displaystyle { \frac {d f} {d u_{horizontal}} = 0 }

which gives

u h o r i z o n t a l = [ L 2 g 2 4 ( 1 + H 2 L 2 ) ] 1 4 u_{horizontal} = \displaystyle { \left [ \frac {L^{2} g^{2}} { 4 \left ( 1 + \frac {H^{2}} {L^{2} } \right ) } \right ]^{\displaystyle {\frac {1}{4}}} }

Thus

u h o r i z o n t a l = 6.620191263 m / s u_{horizontal} = 6.620191263 \ m/s

u v e r t i c a l = 10.711694476 m / s u_{vertical} = 10.711694476 \ m/s

u n e t = 12.59235208 m / s u_{net} = 12.59235208 \ m/s

Reaber John - 7 years, 3 months ago

@Swapnanil Gupta – In my way V horizontal=V vertical.

Swapnanil Gupta - 7 years, 3 months ago

@Swapnanil Gupta – You mean you solved it exactly the way I described as the "correct way" ?

Reaber John - 7 years, 3 months ago

nothing wrong that is the correct answer

Krishna Das - 7 years, 3 months ago

The value of gravitational constant is taken as 9.8. Then tell me, is it sensible to have '6' significant digits in the answer, when only 2 significant digits are used in the solution??? Instead of focusing on a rigid 6 significant digit long answer on a projectile question, the web-site should provide a more flexible and sensible range of answers. I am getting the answer 13 m/s, and i know why it is right... .

Prem Kumar - 7 years, 2 months ago
Sam Thompson
Feb 26, 2014

Let the angle between the horizontal and the angle the ball have value x x , and let the speed of the ball be equal to s s (I can't see the diagram on my laptop, so I don't know if the angle has already been defined). Since there are no forces acting on the horizontal, the speed of the ball relative to the horizontal is always s cos ⁡ x s\cos{x} . Thus, the time required for the ball to pass over the wall is 10 s cos ⁡ x \frac{10}{s\cos{x}} . The speed of the ball relative to the vertical is s sin ⁡ x − 9.8 t s\sin{x}-9.8t . To find the height of the ball at time t t , we take the integral of the vertical speed at time t t (the integral of speed is distance). Taking the integral, we get h = t s sin ⁡ x − 4.9 t 2 h=ts\sin{x}-4.9t^{2} . Substituting the value of t t we derived earlier, we get 2 tan ⁡ x − 98 s 2 cos ⁡ 2 x = 1 2\tan{x}-\frac{98}{s^{2}\cos^{2}{x}}=1 after dividing both sides by 5. Solving for s 2 s^2 , we get s 2 = 98 sin ⁡ 2 x + sin ⁡ 2 x − 1 s^{2}=\frac{98}{\sin^{2}{x}+\sin{2x}-1} . Thus it suffices to find the maximum value of the denominator if we want to find the minimum value of s s . Using Wolfram Alpha or taking the derivative, it is easy to find that the maximum value of the denominator is about 0.618 0.618 . Thus, the minimum value of s s is 98 0.618 = 12.5927... \sqrt{\frac{98}{0.618}}=12.5927... , and we're done.

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Pradeep Ch
Mar 27, 2014

in the equation of trajectory , put 'u' term on one side and the rest on the other side and find du/dx where x is the angle of projection. equate to zero to get tanx = (1+√5)/2 and then finally put this value in the trajectory equation and get the value of u. simple!

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Siddharth Shah
Mar 22, 2014

I simply used the equation of trajectory of a parabolic projectile y=xtanø - 1/2 g^2 x^2 by u^2 (cosø)^2 Which on solving gets u^2 = 98 / (2sinø - cosø) cosø since 'u' should be minimum the trigonometric equation should be maximum on differentiating the T.E wrt ø we get tan2ø = -2 Again on solving the equation for tanø we get tanø= 2 or -1 Taking the tanø = 2 and using the triangle method we get sinø = 2/root 5 and cosø = 1/root 5 substituting the values in the original equation we get u = under root 98*5/3 = 12.7 Hence ANS 12.7 PS sorry for no use of LaTex

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