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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!"

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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.

OghmaOsiris
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1 Answers1

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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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