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Analyzing the Niagara Falls Plunge

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At the beginning of this chapter we asked questions about the motion of the steel chamber holding Dave Munday as he plunged into the water after falling 48 m from the top of Niagara Falls. How long did the fall take? That is, t? How fast was the chamber moving when it hit the water? (What is ?) As you will learn in Chapter 3, if no significant air drag is present, objects near the surface of the Earth fall at a constant acceleration of magnitude ax 9.8 m/s2 . Thus, the kinematic equations can be used to calculate the time of fall and the impact speed. Let’s start by defining our coordinate system. We will take the x axis to be a vertical or up–down axis that is aligned with the downward path of the steel chamber. We place the origin at the bottom of the falls and define up to be positive as shown in Fig. 2-17. (Later when considering motions in two and three dimensions, we will often denote vertical axes as y axes and horizontal axes as x axes, but these changes in symbols will not affect the results of calculations.) We know that the value of the vertical displacement is given by x2  x1 (0 m)  (48 m) 48 m and that the velocity is getting larger in magnitude in the downward (negative direction). Since the velocity is downward and the object is speeding up, the vertical acceleration is also downward (in the negative direction). Its component along the axis of motion is given by ax 9.8 m/s2 . Finally, we assume that Dave Munday’s capsule dropped from rest, so v1 x 0 m/s. Thus we can find the time of fall (t t2  t1) using Eq. 2-17. Solving this equation for the time elapsed during the fall (t2  t1) when the initial velocity v1 x is zero gives This is a fast trip indeed! t t2  t1 ?2(x2  x1) ax ?2( 48

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