Chapter 9.3, Problem 10GP

Solution

To find: The horizontal distance travelled by a track and field long jumper who jumps at given angle with given speed.

The horizontal distance travelled by a track and field long jumper is $29.3\text{ft}$

Given data: The track and field long jumper jumps at angle ${20}^{\text{ο}}$ with initial speed. 27 ft/s.

Method/Formula used:

A motion of an object at some angle say $\theta$ with initial speed v has two components, (1) horizontal motion with constant velocity $v\cos \theta$ along x-axis, and (2) vertical motion with velocity $v\sin \theta$ that changes due to gravitational force of the earth.

Calculations:

The track and field long jumper jumps at angle ${20}^{\text{ο}}$ with initial speed. 27 ft/s. The motion of the jumper is shown in Fig. 1.

  

The jumper travels horizontal distance say x with constant velocity $v\cos \theta$ as there is no acceleration in this direction. Therefore, horizontal distance x travelled by the jumper in time T is

  $x=\left(v\cos \theta \right)T$ … (1)

The time of flight T of an object (time the object remains in air) when projected at an angle $\theta$ with horizontal, with initial velocity v is given by

  $T=\frac{2v\sin \theta }{g}$ … (2)

Here, $g$ equals $32\text{ft/}{\text{s}}^{\text{2}}$ acceleration due to gravity.

Substitute $\frac{2v\sin \theta }{g}$ for T in result (1),

  $\left.\begin{array}{c}x=\left(v\cos \theta \right)\left(\frac{2v\sin \theta }{g}\right)\\ =\frac{2v^2\sin \theta \cos \theta }{g}\end{array}\right\}$ … (3)

The jumper jumps at angle $\theta$ equals ${20}^{\text{ο}}$ with initial speed v equals 27 ft/s.

Substitute $27$ for v, ${20}^{\text{ο}}$ for $\theta$ and $32$ for g in result (3),

  $\begin{array}{c}x=\frac{2{\left(27\right)}^2\sin {20}^{\text{ο}}\cos {20}^{\text{ο}}}{32}\\ =29.29\\ \approx 29.3\end{array}$

Therefore, the horizontal distance travelled by a track and field long jumper is $29.3\text{ft}$