P02G2021 – Projectile Launched from Unknown Height
Projectile Launched from Unknown Height
Someone hits a golf ball from an area that is an unknown distance above the ground. As a result, the ball has an initial velocity of $44\ m/s$ at an angle of $28{}^\circ $ above the horizontal. The ball strikes the ground a horizontal distance of $180.0\ m$ form where the ball was hit.
1. How high $h$ above the ground is the area from which the ball was hit?
View answerHide answer
The kinematic equations of the golf ball as it travels through the air are given by
\left\{ \begin{array}{c}
x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \\
\\
\displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+h} \end{array}
\right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c}
v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \\
\\
v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left(\theta \right)\ } \end{array}
\right.The time $t_f$ at which the ball hits the ground is such that we have $\displaystyle{\left\{ \begin{array}{c}
x\left(t_f\right)=R \\
y\left(t_f\right)=0 \end{array}
\right.}$ where $R=180\ m$ denotes the range of the golf ball. Thus, we solve for $t_f$ as follows
\begin{aligned}
x\left(t_f\right)=R\ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t_f=R \\
\\
&\Rightarrow \ \ \ \ \ t_f=\frac{R}{v_0{\mathrm{cos} \left(\theta \right)\ }} \\
\\
&\Rightarrow \ \ \ \ \ t_f=\frac{180}{44{\mathrm{cos} \left(28\right)\ }}\approx 4.63\ s
\end{aligned}and then solve $y\left(t_f\right)=0$ for $h$ as shown below
\begin{aligned}
y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_f+v_0{\mathrm{sin} \left(\theta \right)\ }t_f+h=0 \\
\\
&\Rightarrow \ \ \ \ \ h=\frac{1}{2}gt^2_f-v_0{\mathrm{sin} \left(\theta \right)\ }t_f
\end{aligned}Thus, the height $h$ from which the golf ball was launched is equal to
\boxed{h=\frac{1}{2}gt^2_f-v_0{\mathrm{sin} \left(\theta \right)\ }t_f=\frac{1}{2}\cdot 9.8\cdot {4.63}^2-44{\mathrm{sin} \left(28\right)\ }\cdot 4.63\approx 9.40\ m}2. What is the speed of the ball just before it strikes the ground?
View answerHide answer
The components of the velocity of the golf ball when it hits the ground are equal to
\begin{aligned}
x\left(t_f\right)&=v_0{\mathrm{cos} \left(\theta \right)\ }t_f=44{\mathrm{cos} \left(28\right)\ }\cdot 4.63\approx 38.85\ m/s \\
\\
y\left(t_f\right)&=-gt_f+v_0{\mathrm{sin} \left(\theta \right)\ }=-9.8\cdot 4.63+44{\mathrm{sin} \left(28\right)\ }\approx -27.71\ \ m/s
\end{aligned}The speed of the ball as it hits the ground is therefore equal to
\boxed{v_f=\sqrt{v_x{\left(t_f\right)}^2+v_y{\left(t_f\right)}^2}=\sqrt{{38.85}^2+{\left(-27.71\right)}^2}\approx 46\ \ m/s}