Chapter 3, Problem 93P

(a)

To determine

To Calculate: The range of the golf-ball when it is hit from thane elevated tee.

Answer to Problem 93P

  $R=194\text{ m}$

Explanation of Solution

Given:

Initial velocity of the golf ball, $v_o=45.0\text{ m/s}$

Launching angle, ${\theta }_o=35.0°$

Net vertical distance covered by tee, $y=20.0\text{m}$

Formula used:

  $R_{\text{elevated}}=\frac{v_o{}^2}{g}\sin 2{\theta }_o$

Calculation:

Plug in the given values in the formula:

  $\begin{array}{c}{R}_{\text{elevated}}=\frac{{\left(45.0\right)}^2}{9.81}\sin \left(2\times {35}^o\right)\\ =193.97\\ R_{\text{elevated}}=194\text{ m}\end{array}$

Conclusion:

The range of the golf-ball when it is hit from thane elevated tee is 194 m.

(b)

To determine

To Compute:- Range, $R=\left(1+\sqrt{1-\frac{2gy}{v_o{}^2{\sin }^2{\theta }_o}}\right)\frac{v_o{}^2}{2g}\sin 2{\theta }_o$

Explanation of Solution

Given data:

Horizontal acceleration, $a_x=0{\text{ m/s}}^2$

Vertical acceleration, $a_y=-9.8{\text{ m/s}}^2$

Vertical displacement, $y=-h$

Formula used:

The second equation of kinematics for the constant acceleration in horizontal direction:

  $x=u_{x}t+\frac{1}{2}{a}_x{t}^2\mathrm{...}\text{(i)}$

The second equation of kinematics for the constant acceleration in vertical direction:

  $y=u_{y}t+\frac{1}{2}{a}_y{t}^2\mathrm{...}\text{(ii)}$

Calculation:

Vertical component of the velocity, $u_y=v_o\sin {\theta }_o$

Horizontal component of the velocity, $u_x=v_o\cos {\theta }_o$

Plug in the values in equation (i):

  $\begin{array}{l}x=v_o\cos {\theta }_o\times t+\frac{1}{2}\times 0\times t^2\\ x=v_o\cos {\theta }_o\times t\\ \therefore t=\frac{x}{v_o\cos {\theta }_o}\end{array}$

The second equation of kinematics for the projectile’s motion:

  $y=h+u_{y}t+\frac{1}{2}{a}_y{t}^2$

Plugging in the values:

  $\begin{array}{c}y=h+u_{y}t+\frac{1}{2}{a}_y{t}^2\\ =h+v_o\sin {\theta }_o\times \frac{x}{v_o\cos {\theta }_o}+\frac{1}{2}\times \left(-g\right)\times {\left(\frac{x}{v_o\cos {\theta }_o}\right)}^2\\ =h+x\left(\tan {\theta }_o\right)-\frac{1}{2}g{\left(\frac{x}{v_o\cos {\theta }_o}\right)}^2\end{array}$

When projectile hits the ground: $x=R,y=0$

Therefore,

  $\begin{array}{l}h+R\left(\tan {\theta }_o\right)-\frac{1}{2}g{\left(\frac{R}{v_o\cos {\theta }_o}\right)}^2=0\\ \Rightarrow 2h{\left(v_o\cos {\theta }_o\right)}^2+2R\left(\tan {\theta }_o\right){\left(v_o\cos {\theta }_o\right)}^2-g{R}^2=0\\ \Rightarrow 2h{v}_o{}^2{\cos }^2{\theta }_o+2R\times \frac{\sin {\theta }_o}{\cos {\theta }_o}\times v_o{}^2{\cos }^2{\theta }_o-g{R}^2=0\\ \Rightarrow g{R}^2-\left(2v_o{}^2\sin {\theta }_o\cos {\theta }_o\right)R-2h{v}_o{}^2{\cos }^2{\theta }_o=0\end{array}$

Solving for Rby quadratic equation:

  $\begin{array}{c}R=\frac{-\left(-2v_o{}^2\sin {\theta }_o\cos {\theta }_o\right)\pm \sqrt{{\left(-2v_o{}^2\sin {\theta }_o\cos {\theta }_o\right)}^2-4\times g\times \left(-2h{v}_o{}^2{\cos }^2{\theta }_o\right)}}{2\times g}\\ =\frac{2v_o{}^2\sin {\theta }_o\cos {\theta }_o\pm 2v_o{}^2\cos {\theta }_o\sqrt{{\sin }^2{\theta }_o+\frac{2gh}{v_o{}^2}}}{2g}\\ =\frac{2v_o{}^2\sin {\theta }_o\cos {\theta }_o\pm 2v_o{}^2\cos {\theta }_o\sin {\theta }_o\sqrt{1+\frac{2gh}{v_o{}^2{\sin }^2{\theta }_o}}}{2g}\\ =\frac{\left(2v_o{}^2\sin {\theta }_o\cos {\theta }_o\right)}{2g}\left(1+\sqrt{1+\frac{2gh}{v_o{}^2{\sin }^2{\theta }_o}}\right)\\ =\left(1+\sqrt{1+\frac{2gh}{v_o{}^2{\sin }^2{\theta }_o}}\right)\frac{v_o{}^2}{2g}\sin 2{\theta }_o\end{array}$

Plugging in $h=-y$

  $R=\left(1+\sqrt{1-\frac{2gy}{v_o{}^2{\sin }^2{\theta }_o}}\right)\frac{v_o{}^2}{2g}\sin 2{\theta }_o$

Conclusion:

  $R=\left(1+\sqrt{1-\frac{2gy}{v_o{}^2{\sin }^2{\theta }_o}}\right)\frac{v_o{}^2}{2g}\sin 2{\theta }_o$

(c)

To determine

The numerical value for the range and the percentage error.

Answer to Problem 93P

  $R=219\text{ m}$

Percentage error $=11\%$

Explanation of Solution

Given data:

Initial velocity of the golf ball, $v_o=45.0\text{ m/s}$

Launch angle, ${\theta }_o=35.0°$

Acceleration due to gravity, g = 9.81 m/s²

  $y=-20.0\text{ m}$

Formula used:

  $R=\left(1+\sqrt{1-\frac{2gy}{v_o{}^2{\sin }^2{\theta }_o}}\right)\frac{v_o{}^2}{2g}\sin 2{\theta }_o$

Calculation:

Plugging in the given values in the formula:

  $\begin{array}{c}R=\left(1+\sqrt{1-\frac{2\times 9.81\times \left(-20\right)}{{\left(45.0\right)}^2{\sin }^2\left({35}^o\right)}}\right)\frac{{\left(45.0\right)}^2}{2\times 9.8}\sin \left(2\times {35}^o\right)\\ \approx 219\text{ m}\end{array}$

The percentage error:

  $\begin{array}{c}=\left|\frac{R-R_{elevated}}{R}\right|\\ =\left|\frac{219-194}{219}\right|\\ \approx 11\%\end{array}$

Conclusion:

The range: $219\text{ m}$

The percentage error: $11\%$