Chapter 6.1, Problem 10SSC

To determine

The maximum height attained by the ball.

Answer to Problem 10SSC

  $3.62\text{m}$

Explanation of Solution

Given:

A softball is tossed in the air with an initial velocity of 11m/s at an angle of $50°$ above 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 displacement of 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 are:

  $\begin{array}{l}{v}_{ix}=v_i\cos 50°=11\cos 50°=7.07\text{m/s}\\ v_{iy}=v_i\sin 50°=11\sin 50°=8.42\text{m/s}\end{array}$

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

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

Negative sign shows that this acceleration acts in opposite direction (downward direction) to the motion of the ball.

Also, it is known that the ball comes to momentarily rest at the maximum height, thus at maximum height,

  $v_{fy}=0\text{m/s}$

Now, considering the vertical motion of the ball to determine the time taken to attain maximum height as,

  $\begin{array}{c}{v}_{fy}=v_{iy}+a_{y}t\\ 0=8.42+\left(-9.8\right)t\\ t=0.859\text{s}\end{array}$

Then, the maximum height attained by the ball will be,

  $\begin{array}{c}{d}_{\max }=v_{iy}t+\frac{1}{2}{a}_y{t}^2\\ =\left(8.42\right)\left(0.859\right)+\frac{1}{2}\left(-9.8\right){\left(0.859\right)}^2\\ =7.232-3.615\\ =3.62\text{m}\end{array}$

Conclusion:

Thus, the maximum height attained by the ball is $3.62\text{m}$ .