Question 4.2: The Long Jump A long jumper (Fig. 4.12) leaves the ground at...
The Long Jump
A long jumper (Fig. 4.12) leaves the ground at an angle of 20.0° above the horizontal and at a speed of 11.0 m/s.
(A) How far does he jump in the horizontal direction?
(B) What is the maximum height reached?
(A) Conceptualize The arms and legs of a long jumper move in a complicated way, but we will ignore this motion. We model the long jumper as a particle and conceptualize his motion as equivalent to that of a simple projectile.
Categorize We categorize this example as a projectile motion problem. Because the initial speed and launch angle are given and because the final height is the same as the initial height, we further categorize this problem as satisfying the conditions for which Equations 4.19 and 4.20 can be used. This approach is the most direct way to analyze this problem, although the general methods that have been described will always give the correct answer.
h =\frac{v_i{}^2 \sin ^2 \theta_i}{2 g} (4.19)
R =\frac{v_i{}^2 \sin 2 \theta_i}{g} (4.20)
Analyze
Use Equation 4.20 to find the range of the jumper:
R =\frac{v_i{}^2 \sin 2 \theta_i}{g}=\frac{(11.0 m/s)^2 \sin 2\left(20.0^{\circ}\right)}{9.80 m/s^2}=7.94 m(B) Analyze
Find the maximum height reached by using Equation 4.19:
h= \frac{v_i{ }^2 \sin ^2 \theta_i}{2 g}=\frac{(11.0 m/s)^2\left(\sin 20.0^{\circ}\right)^2}{2\left(9.80 m/s^2\right)}=0.722 mFinalize Find the answers to parts (A) and (B) using the general method. The results should agree. Treating the long jumper as a particle is an oversimplification. Nevertheless, the values obtained are consistent with experience in sports. We can model a complicated system such as a long jumper as a particle and still obtain reasonable results.