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Chapter 12: Problem 14

In the 1996 Olympic Games, the Bulgarian high jumper Stefka Kostadinova set a women's Olympic record for this event with a jump of \(2.05 \mathrm{~m}\); see Fig. \(12-17\). Other things being equal, how high might she have jumped on the Moon, where the surface gravity is only \(1.67 \mathrm{~m} / \mathrm{s}^{2} ?\) (Hint: The height that "counts" is the vertical distance her center of gravity rose after her feet left the ground. Assume that, at the instant her feet lost contact, her center of gravity was \(110 \mathrm{~cm}\) above ground level. Assume also that, as she clears the bar, her center of gravity is at the same height as the bar.)

Short Answer

Expert verified
The answer is that Stefka Kostadinova might have jumped approximately \(11.31 \, m\) high on the Moon.

Step by step solution

01

Identify the problem type

Here, we are dealing with a physics problem that involves free fall under gravity. Specifically, we are asked to compare heights of free fall on Earth and on the Moon.
02

Calculate the total time for ascent and descent on Earth

The equation for height in a physics motion problem is \( h = h_0 + v_0t - 0.5gt^2 \). However, since on the highest point the velocity is 0, the equation can be rewritten in terms of time \( t = \sqrt{2h/g} \). When considering the time for ascent and descent, the time actually doubles. In our case, the initial height \( h_0 \) is 1.10m and the height gained \( h \) is 0.95m(2.05m - 1.10m). So, for the Earth where gravity \( g=9.8 \, m/s^2 \), the total time becomes \( t = 2\sqrt{2*0.95/9.8} \).
03

Calculate the total height she might have jumped on the Moon

We can calculate the total height she might have jumped on the Moon by taking into account that ascent and descent times are equal to that of Earth, but gravity is different. Re-writing our equation from step 2 and solve for height: \( h = g*(t^2)/2 \). Putting the calculated value for \( t \) and using moon's gravity \( g=1.67 \, m/s^2 \), we get the height she could have jumped on the Moon avoiding the initial height. To get the total height, we just need to add the initial height which is \( 1.10 \, m \) on Moon too.
04

Verify the results

Check the results for plausibility by considering that the gravity on the Moon is significantly smaller than that on Earth. Therefore, the height jumped should be significantly higher.

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Most popular questions from this chapter

In one dimension, the magnitude of the gravitational force of attraction between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$F_{x}(x)=G \frac{m_{1} m_{2}}{x^{2}}$$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty,(b)\) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=\) \(x_{1}+d ?\)

The potential energy corresponding to a certain two-dimensional force is given by \(U(x, y)=\frac{1}{2} k\left(x^{2}+y^{2}\right) .(a)\) Derive \(F_{x}\) and \(F_{y}\) and describe the vector force at each point in terms of its coordinates \(x\) and \(y .(b)\) Derive \(F_{r}\) and \(F_{\theta}\) and describe the vector force at each point in terms of the polar coordinates \(r\) and \(\theta\) of the point. ( \(c\) ) Can you think of a physical model of such a force?

A body is rolling horizontally without slipping with speed \(v\). It then rolls up a hill to a maximum height \(h\). If \(h=3 v^{2} / 4 g\) what might the body be?

A particle moves along the \(x\) axis under the influence of a conservative force that is described by $$\overrightarrow{\mathbf{F}}=-\alpha x e^{-\beta x^{2}} \hat{\mathbf{i}}$$ where \(\alpha\) and \(\beta\) are constants. Find the potential energy function \(U(x)\).

An object falls from rest from a height \(h\). Determine the kinetic energy and the potential energy of the object as a function \((a)\) of time and \((b)\) of height. Graph the expressions and show that their sum - the total mechanical energy - is constant in each case.
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