I saw this picture on one of my social media sites with the caption, "I'd do this in a heart beat! Who's with me!"

enter image description here

I was about to go balls to the walls and say, "I'm in! When and where??" But then I got to thinking, how fast would I be going when I hit the water? If I were going too fast, would it hurt me?

SO I was trying to figure this out, and I'm not very good at physics so I was wondering if you guys could help me out.

I estimate the guy is 90 kg in mass, the wire is angled pi/6 from the horizontal and he's about 50 meters above the water when he starts (all estimates...).

What is the formulas I need to figure out the speed the guy will be going once he hits the water? I know there's some calculus in there, and I'm pretty good at calculus.

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    $\begingroup$ It just looks like he's 150 feet above the water, so I was just estimating. $\endgroup$ – OghmaOsiris May 9 '13 at 3:36
  • $\begingroup$ Sorry wrong estimate :), it would be about 100Km/hr $\endgroup$ – ABC May 9 '13 at 3:39

The only force which works is gravity$^1$. So, change in gravitational potential energy equals final Kinetic energy(assume initial is zero). $$mgh=mv^2/2$$ $$v=\sqrt{2gh}$$

here $h$ is vertical height traversed.See the velocity does not depend on angle of string, mass of body too..

Let's see the kinematics of body.

The length of string is $h cosec\theta$ ($\theta $ being angle with horizontal assumed $\pi/6$)

acceleration of body along the string=$g\sin\theta$

Now $\text{using} : v^2=u^2+2as$

$$v^2=0+2\times h cosec\theta\times g \sin\theta$$ $$v=\sqrt{2gh}$$

Working in differentials

for $v$ along the rope. $$dv/dt=v\dfrac{dv}{dx}=a$$ $$\int_0^{v_f} v.dv=\int_0^{hcosec\theta} a.dx=ax\Bigg|_0^{hsosec\theta}$$ $$\dfrac{v_f^2}2=gsin\theta.hcosec\theta \ \ ; \ \ a=gsin\theta$$

$1)$Assuming the pulley being used to slide to be friction less.Though not possible.Also the rope is assumed to be in-extensible and straight.

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    $\begingroup$ Would the acceleration due to gravity be the same when going at an angle, though? I thought it would be a lot less than 9.81 m/s $\endgroup$ – OghmaOsiris May 9 '13 at 3:41
  • $\begingroup$ @OghmaOsiris See I worked out energies , the gravitational potential energy just change in vertical displacement.So, the work done is just $mgh_{\text{vertical}}$ $\endgroup$ – ABC May 9 '13 at 3:43
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    $\begingroup$ I was hoping there woul be some integration involved :(... I really like calculus lol. $\endgroup$ – OghmaOsiris May 9 '13 at 4:05
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    $\begingroup$ @007 It looks like you integrated the velocity anyways. And I understand things better when put into the reference frame of math. The way you explained it isn't intuitive at all to me. It seemed more intuitive to integrate the velocity along a path. $\endgroup$ – OghmaOsiris May 9 '13 at 4:37
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    $\begingroup$ As an aside to physiology, be careful about hitting the water at that speed in that position. You could wind up forcing water to go where it really shouldn't... and can't... $\endgroup$ – DJohnM May 9 '13 at 6:32

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