Chapter 10, Problem 9AP

A badminton shuttlecock is struck by a racket at a 35° angle, giving it an initial speed of 10 m/s. How high will it go? How far will it travel horizontally before being contacted by the opponent’s racket at the same height from which it was projected? (Answer: d_v = 1.68 m; d_h = 9.58 m)

Summary Introduction

To determine: The horizontal and vertical distance of the badminton shuttlecock.

Answer to Problem 9AP

Vertical distance travelled by the shuttle cock is $1.68\text{m}$ and horizontal distance travelled by the shuttle cock is $9.58\text{m}$.

Explanation of Solution

Calculation:

Write the relation to find the vertical distance travelled by the ball.

$v_2^2={\left(v_1\sin \theta \right)}^2+2a{d}_v$

Here, $v_2$ is the final velocity of the ball, $v_1$ is the initial velocity of the ball, $\theta$ is the angle struck by the racquet, $a$ is the acceleration due to gravity, and $d_v$ is the vertical distance.

Substitute $10\text{m}/\text{s}$ for $v_1$, $0\text{m}/\text{s}$ for $v_2$, $35°$ for $\theta$, and $-9.81{\text{m}/\text{s}}^2$ for $a$ to find $d_v$.

$\begin{array}{c}0\text{m}/\text{s}={\left(\left(10\text{m}/\text{s}\right)\sin 35°\right)}^2+2\left(-9.81\text{m}/{\text{s}}^2\right)d_v\\ d_v=\frac{32.89902{{\text{m}}^2/\text{s}}^2}{19.62\text{m}/{\text{s}}^2}\\ =1.67681\text{m}\\ \approx 1.68\text{m}\end{array}$

Thus, vertical distance travelled by the shuttle cock is $1.68\text{m}$.

Write the relation to find the time of flight.

$v_2=v_1\sin \theta +at$

Here, $t$ is the time of flight.

Substitute $10\text{m}/\text{s}$ for $v_1$, $0\text{m}/\text{s}$ for $v_2$, $35°$ for $\theta$, and $-9.81{\text{m}/\text{s}}^2$ for $a$ to find $t$.

$\begin{array}{c}0\text{m}/\text{s}=\left(10\text{m}/\text{s}\right)\sin 35°+\left(-9.81\text{m}/{\text{s}}^2\right)t\\ t=\frac{5.735\text{m}/\text{s}}{9.81\text{m}/{\text{s}}^2}\\ =0.584686\text{s}\\ \approx 0.585\text{s}\end{array}$

$\begin{array}{c}\text{Total}\text{time}\text{of}\text{flight}=2t\\ =2\times 0.585\text{s}\\ =\text{1}\text{.169}\text{s}\\ \approx \text{1}\text{.17}\text{s}\end{array}$

Thus, total time of flight is $1.17\text{s}$.

Write the relation to find the horizontal distance travelled by the ball.

$d_h=\left(v_1\cos \theta \right)t+\frac{1}{2}a{t}^2$

Here, $d_h$ is the horizontal distance.

Substitute $10\text{m}/\text{s}$ for $v_1$, $35°$ for $\theta$, and $1.17\text{s}$ for t to find $d_h$.

$\begin{array}{c}{d}_h=\left(10\text{m}/\text{s}\cos 35°\right)\left(1.17\text{s}\right)\\ =9.578928\text{m}\\ \approx 9.58\text{m}\end{array}$

Thus, horizontal distance travelled by the shuttle cock is $9.58\text{m}$.

Conclusion

Therefore, vertical distance travelled by the shuttle cock is $1.68\text{m}$ and horizontal distance travelled by the shuttle cock is $9.58\text{m}$.