Chapter 6, Problem 1STP

To determine

To Choose: The correct option.

Answer to Problem 1STP

Option C. $8.90\text{m}$

Explanation of Solution

Given info:

Initial height from which ball is thrown, $h=1.60\text{}\text{m}$

The angle at which the ball is thrown, $\theta =41.0°$

Initial velocity with which the ball is thrown, $v_i=9.40\text{m/s}$

Formula used:

First equation of motion:

  $v_f=v_i+at$

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

And,

The distance traveled by an object with a constant acceleration is given by,

  $d_f=v_{i}t+\frac{1}{2}a{t}^2$

And,

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,

  $\begin{array}{c}{v}_{ix}=v_i\cos 41°\\ =9.4\cos 41°\\ =7.09\text{m/s}\\ v_{iy}=v_i\sin 41°\\ =9.4\sin 41°\\ =6.17\text{m/s}\end{array}$

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 is used since the acceleration acts in downward direction and opposes the motion of the ball

Now, considering the vertical motion of the ball to determine the time taken for the projectile motion as,

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

In above equation, $d_{fy}=0\text{m}$ as the ball returns to the same height.

Then, using given and calculated values in above,

  $\begin{array}{c}0=\left(6.17\right)t+\frac{1}{2}\left(-9.8\right)t^2\\ 0=t\left(6.17-4.9t\right)\end{array}$

Considering only the positive value of “t”,

  $t=1.259\text{s}$

Now, since no acceleration acts on ball in horizontal direction, hence horizontal distance covered by the ball in $1.26\text{}\text{s}$ is,

  $\begin{array}{c}{d}_{fx}=v_{ix}\times t\\ =7.09\times 1.259\\ \approx 8.90\text{m}\end{array}$

Conclusion:

Thus, the horizontal distance covered by the ball is $8.90\text{m}$ .