Chapter 6.1, Problem 9SSC

To determine

The distance covered by the ball horizontally before it hits the ground.

Answer to Problem 9SSC

  $27\text{m}$

Explanation of Solution

Given:

A tennis ball is thrown out of a window, which is 28m above the ground, at an initial velocity of 15m/s and $20°$ below the horizontal.

Formula used:

Final velocity with constant acceleration can be obtained by first equation of motion:

  $v_f=v_i+at$

Where, $v_f$ is the final velocity, $v_i$ is the initial velocity and $t$ is the time taken.

The displacementof an object with a constant acceleration can be obtained by third equation of motion:

  $d_f=v_{i}t+\frac{1}{2}a{t}^2$

Where, $d_f$ is displacement, $v_i$ is the initial velocity, a is the acceleration and $t$ is the time taken

The rectangular components of a vector A making an angle of $\theta$ with the horizontal axis or x -axis, is given by:

  $A_x=A\cos \theta$

  $A_y=A\sin \theta$

Calculation:

The components of the initial velocity in the vertical and horizontal direction is,

  $\begin{array}{l}{v}_{ix}=v_i\cos 20°=15\cos 20°=14.09\text{m/s}\\ v_{iy}=v_i\sin 20°=15\sin 20°=5.13\text{m/s}\end{array}$

Acceleration acting on the ball is in the vertical direction which is due to gravity only, hence it can be written as

  $\begin{array}{l}{a}_x=0\mathrm{m}/s^2\\ a_y=9.8\mathrm{m}/s^2\end{array}$

Now, consider the vertical motion of the ball to determine the time taken for the projectile motion.

  $d_{fy}=v_{iy}t+\frac{1}{2}{a}_y{t}^2$

Using given and calculated values in above,

  $\begin{array}{l}28=5.13t+\frac{1}{2}\left(9.8\right)t^2\\ 4.9t^2+5.13t-28=0\end{array}$

Considering only the positive value of t :

  $t=1.92\text{s}$

Now, since no acceleration acts on ball in horizontal direction, hence horizontal distance covered by the ball in 2.044s is,

  $\begin{array}{c}{d}_{fx}=v_{ix}\times t\\ =14.09\times 1.92\\ =27\text{m}\end{array}$

Conclusion:

Thus, the horizontal distance covered by the ball is $27\text{m}$ .