Chapter 6, Problem 46A

To determine

Distance of player from the board.

Answer to Problem 46A

  $3.174\text{m}$

Explanation of Solution

Given:

Initial horizontal velocity of dart is $u_x=12.4\text{m/s}$

Initial vertical velocity of dart is $u_y=0\text{m/s}$

Vertical displacement, $y=0.32\text{m}$

Formula Used:

Second equation of motion:

  $s=ut+\frac{1}{2}a{t}^2$

Where, s is the displacement, t is the time, a is the acceleration and u is the initial velocity.

Calculation:

For vertical motion:

  $y=u_{y}t-\frac{1}{2}g{t}^2$ ….. (1)

In horizontal motion:

  $x=u_{x}t$ ….. (2)

As initial vertical velocity of dart is zero $u_y=0\text{m/s}$

So, from (1)

  $\begin{array}{c}t=\sqrt{\frac{-2y}{g}}\\ =\sqrt{\frac{-2\times \left(-0.32\right)}{9.8}}\\ =0.256\text{s}\end{array}$

Now, horizontal distance is given by equation (2)

  $\begin{array}{c}x=12.4\times 0.256\\ =3.174\text{m}\end{array}$

Conclusion:

So, distance of player from the board is $3.174\text{m}$ .