Footballer range and maximum height

Problem

A footballer kicks a ball with an initial velocity of $20\;\text{m/s}$ at an angle of $30^\circ$ to the ground. Calculate: (a) the maximum height reached; (b) the range; (c) the flight time. Neglect air resistance.

Solution

Strategy: decompose the motion. Projectile motion is treated by splitting the initial velocity into its two Cartesian components: the horizontal one stays constant (no horizontal force), the vertical one is subject to the acceleration of gravity $g \approx 9{,}81\;\text{m/s}^2$ downwards. With $v_0 = 20\;\text{m/s}$ and $\theta = 30^\circ$:

$v_{0x} = v_0 \cos\theta = 20\cos 30^\circ \approx 17{,}3\;\text{m/s}, \qquad v_{0y} = v_0 \sin\theta = 20\sin 30^\circ = 10{,}0\;\text{m/s}$

(a) Maximum height. At the highest point the vertical velocity component vanishes. From $v_y^2 = v_{0y}^2 - 2 g\, h_{\max}$ with $v_y = 0$:

$h_{\max} = \frac{v_{0y}^2}{2g} = \frac{(10{,}0)^2}{2\cdot 9{,}81} \approx 5{,}1\;\text{m}$

$\ev{h_{\max} \approx 5{,}1\;\mathrm{m}}$

(c) Flight time. It is convenient to work this out before the range. The ball leaves and lands at the same height; the flight time is twice the rise time $t_{\text{sal}} = v_{0y}/g$:

$t_{\text{volo}} = \frac{2 v_{0y}}{g} = \frac{2\cdot 10{,}0}{9{,}81} \approx 2{,}0\;\text{s}$

$\ev{t_{\text{volo}} \approx 2{,}0\;\mathrm{s}}$

(b) Range. Along the horizontal the motion is uniform at velocity $v_{0x}$, for the whole flight time:

$G = v_{0x}\, t_{\text{volo}} = 17{,}3\cdot 2{,}0 \approx 35{,}3\;\text{m}$

$\ev{G \approx 35{,}3\;\mathrm{m}}$

Check. The range can also be obtained from the compact formula $G = v_0^2 \sin(2\theta)/g = (20)^2 \sin 60^\circ / 9{,}81 \approx 35{,}3\;\text{m}$, in agreement with the component-wise calculation.

Collegamenti

Argomenti: Kinematics · Vectors Concetti: Projectile motion Competenze: Vector decomposition · Use of trigonometry Metodi: Decomposition into Cartesian components