# Morbid Curiosity

On Sunday I headed to Toronto with several friends to complete the CN Tower Edge Walk adventure. The idea: walk the outside of the CN Tower above the revolving restaurant (a height of 356 metres – or 1168 feet) above the streets of Toronto. For safety reasons, we were obviously tethered to the building. The adventure was fantastic, the weather couldn’t have been more perfect, and the experience was beyond compare.

Of course, given that the group of us are all in some way academics, conversations about the probability of falling (based on not one, but two tethers snapping), how long it would take to hit the ground, and what object one might want to throw from the tower if they could (and assuming no one below would be hurt) made up a good portion of our adventure.

But this post is not about the object I would want to throw1. Instead, it’s about how long it would take for me to hit the ground in the extremely unlikely event that I slipped after both tethers broke or the very sturdy supports bolted into the side of the tower just happened to fail.

For this calculation I’m going to ignore air resistance. I’m also going to assume that I miss the lower level of the tower that juts out slightly from where the Edge Walk occurs – hence, I’m ignoring the possibility of being slowed or stopped, or experiencing any unfortunate bounces along the way. I’m also going to assume that I didn’t jump, so my initial velocity would be 0.

Clearly this is a simple calculation that most of us would have encountered in high school or first year Calculus (or possibly Physics). Recall that the distance an object travels can be written $d=\displaystyle{\frac{g\cdot t^{2}}{2}}$, where $g$ is gravity ($9.8\frac{m}{s^{2}}$ down), and $t$ is time. We want to solve for the time $t$ it takes to travel from the Edge Walk platform to the ground below – 356 metres. Solving for $t$ we find that $t=\displaystyle{\sqrt{\frac{2d}{g}}}=\displaystyle{\sqrt{\frac{2\cdot 356}{9.8}}}\approx 8.5$ seconds.

To calculate the speed I would hit the ground, we simply plug the total travel time into the standard velocity equation $v=g\cdot t$. Assuming no air resistance, our travel time of $t\approx 8.5$ seconds gives a velocity of $v=9.8\cdot 8.5=83.3$ metres per second (or approximately 299.88 kilometres per hour).

So there we have it. Assuming negligible air resistance and perfect conditions, it would take 8.5 seconds for me to hit the pavement, and I’d do so at a speed of almost 300 km/hr.

Fortunately for me, I didn’t slip and so I have no empirical evidence to support this claim. 🙂

1 For those curious, I would want to throw a rubber bouncy ball off of the tower.