Chapter 3, Problem 32P

To determine

The horizontal distance travelled by the shot.

Answer to Problem 32P

The horizontal distance travelled by the shot is 25.8m.

Explanation of Solution

Given:

The initial speed is 15.5 m/s and angle of projection is $34°$ .

The height above the ground( h ) is 2.20m.

Calculation:

Evaluating the vertical and the horizontal component of the velocity:

  $\begin{array}{l}{v}_{xo}=v_o\cos \theta \\ =15.5\cos {34}^o\\ =12.9\text{m/s}\\ v_{yo}=v_o\sin \theta \\ =15.5\sin {34}^o\\ =8.67\text{m/s}\end{array}$

Now, use the positive direction as up and the origin at the release point, the time required for fall from the virtual motion is expressed as:

  $\begin{array}{l}y=y_o+v_{oy}t+\frac{1}{2}{a}_y{t}^2\\ -2.2=0+8.67t+\frac{1}{2}\left(-9.8\right)t^2\\ 4.9t^2-8.67t-2.2=0\end{array}$

Solving the above equation:

  $t=1.99\text{s,-0}\text{.228s}$

Since, the negative value of the time is not possible, hence the time will be:

  $t=2\text{s}$

Now, evaluating the horizontal distance:

  $\begin{array}{l}x=x_o+v_{ox}t\\ =0+12.9\left(2\right)\\ =25.8\text{m}\end{array}$