Chapter 3, Problem 43P

(a)

To determine

The initial velocity that must be imparted to the golf ball so that it crosses over a tree and lands on the green.

Answer to Problem 43P

The initial velocity of the golf ball is found to have a magnitude $17.5\text{ m/s}$ and is at an angle ${52.7}^{\text{o}}$ to the horizontal.

Explanation of Solution

Given:

Horizontal distance to the tree, $x=15.0\text{ m}$

Height of the tree, $y=9.8\text{ m}$

The golf ball just grazes over the top of the tree, hence its maximum height in its trajectory is equal to the height of the tree.

Formula used:

For the vertical motion of the ball,

  $v_y^2=v_{0y}^2-2gy$.....(1)

  $v_y=v_{0y}-gt$.....(2)

Here, $v_y$ is the vertical component of the ball’s initial velocity $v_0$ at a vertical height y ,at a time t , $v_{0y}$ is the initial vertical component of $v_0$ at the time of projection and g is the acceleration of free fall.

The horizontal motion of the ball is uniform, since no force acts along the horizontal direction.

For horizontal motion

  $\begin{array}{l}{v}_x=v_{0x}\\ v_x=\frac{x}{t}\end{array}$.....(3)

Here, $v_{0x}$ is the horizontal component of the velocity $v_0$ at the time of projection and $v_x$ is its horizontal velocity at time t after it travels a horizontal distance x .

The magnitude of the velocity of projection is given by,

  $v_0=\sqrt{v_{0x}^2+v_{0y}^2}$.....(4)

The angle of projection is given by

  $\theta ={\tan }^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)$.....(5)

Calculation:

The tree is at the topmost point of the ball’s trajectory. At this point, the vertical component of its velocity becomes equal to zero.

Substitute $\left(0\text{ m/s}\right)$ for $v_y$ , $9.8{\text{ m/s}}^2$ for g , $\left(9.8\text{ m}\right)$ for y and calculate the value of $v_{0y}$ .

  $\begin{array}{l}{v}_y^2=v_{0y}^2-2gy\\ {\left(0\text{ m/s}\right)}^2=v_{0y}^2-2\left(9.8{\text{ m/s}}^2\right)\left(9.8\text{ m}\right)\\ v_{0y}=\sqrt{2\left(9.8{\text{ m/s}}^2\right)\left(9.8\text{ m}\right)}\\ =13.9\text{ m/s}\end{array}$

Calculate the time taken to reach the height t using equation (2) and substituting $\left(0\text{ m/s}\right)$ for $v_y$

  $9.8{\text{ m/s}}^2$ forg and $\left(13.9\text{ m/s}\right)$ for $v_{0y}$ .

  $\begin{array}{l}{v}_y=v_{0y}-gt\\ \left(0\text{ m/s}\right)=\left(13.9\text{ m/s}\right)-\left(9.8{\text{ m/s}}^2\right)t\\ t=\frac{\left(13.9\text{ m/s}\right)}{\left(9.8{\text{ m/s}}^2\right)}\\ =1.42\text{ s}\end{array}$

In the time t the ball travels a horizontal distance x . Substitute $15.0\text{ m}$ for x and $1.42\text{ s}$ for t in equation (3).

  $\begin{array}{c}{v}_{0x}=\frac{x}{t}\\ =\frac{15.0\text{ m}}{1.42\text{ s}}\\ =10.6\text{ m/s}\end{array}$

Use the calculated values of $v_{0x}$ and $v_{0y}$ in equations (4) and (5) to calculate the magnitude and direction of $v_0$ .

  $\begin{array}{c}{v}_0=\sqrt{v_{0x}^2+v_{0y}^2}\\ =\sqrt{{\left(10.6\text{ m/s}\right)}^2+{\left(13.9\text{ m/s}\right)}^2}\\ =17.5\text{ m/s}\end{array}$

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)\\ ={\tan }^{-1}\left(\frac{13.9\text{ m/s}}{10.6\text{ m/s}}\right)\\ =52.7°\end{array}$

Conclusion:

Thus the initial velocity that must be imparted to the golf ball so that it crosses over a tree and lands on the green is found to have magnitude $17.5\text{ m/s}$ and is at an angle ${52.7}^{\text{o}}$ to the horizontal.

(b)

To determine

The horizontal distance the ball travels after crossing the tree before hitting the ground.

Answer to Problem 43P

The horizontal distance the ball travels after crossing the tree before hitting the ground is found to be $10.6\text{ m}.$

Explanation of Solution

Given:

The vertical height of the ground from the point of launch, $y_1=4.9\text{ m}$

The vertical component of the ball’s initial velocity, $v_{0y}=13.9\text{ m/s}$

The horizontal component of the ball’s initial velocity, $v_{0x}=10.6\text{ m/s}$

Time taken by the ball to cross the tree, $t=1.42\text{ s}$

Formula used:

If the time of flight of the ball is $t_1$ , then

  $y_1=v_{0y}{t}_1-\frac{1}{2}g{t}_1^2$.....(6)

The time taken to travel the distance d after crossing the tree is given by

  $\Delta t=t_1-t$.....(7)

The horizontal distance d is given by

  $d=v_{0x}\Delta t$.....(8)

Calculation:

Substitute the given values of variables in equation (6) to calculate the value of $t_1$ .

  $\begin{array}{l}{y}_1=v_{0y}{t}_1-\frac{1}{2}g{t}_1^2\\ \left(4.9\text{ m}\right)=\left(13.9\text{ m/s}\right)t_1-\frac{1}{2}\left(9.8{\text{ m/s}}^2\right)t_1^2\end{array}$

Simplify the equation:

  $\left(4.9{\text{ m/s}}^2\right)t_1^2-\left(13.9\text{ m/s}\right)t_1+\left(4.9\text{ m}\right)=0$

Solve the quadratic to obtain the values of $t_1$ .

  $\begin{array}{c}{t}_1=\frac{-\left(-13.9\text{ m/s}\right)\pm \sqrt{{\left(-13.9\text{ m/s}\right)}^2-4\left(4.9{\text{ m/s}}^2\right)\left(4.9\text{ m}\right)}}{2\left(4.9{\text{ m/s}}^2\right)}\\ t_1=0.412\text{ s}\\ t_1=2.42\text{ s}\end{array}$

Since, it is required to find the time taken to reach the green, $t_1>t$ . Hence,

  $t_1=2.42\text{ s}$

Calculate the value of $\Delta t$ using equation (7).

  $\begin{array}{c}\Delta t=t_1-t\\ =\left(2.42\text{ s}\right)-\left(1.42\text{ s}\right)\\ =1.00\text{ s}\end{array}$

Calculate the distance d using equation (8).

  $\begin{array}{c}d=v_{0x}\Delta t\\ =\left(10.6\text{ m/s}\right)\left(1.00\text{ s}\right)\\ =10.6\text{ m}\end{array}$

Conclusion:

Thus, the horizontal distance the ball travels after crossing the tree before hitting the ground is found to be $10.6\text{ m}.$