91Ó°ÊÓ

As the ball is thrown horizontally from the top of the building, so the initial vertical component of the velocity is zero.

If\({v_x}\)is the horizontal component of the velocity,\({v_y}\)is the horizontal component of the velocity, then the resultant of the velocity is given by,

\({v_R} = \sqrt {v_x^2 + v_y^2} \)

Here\({v_R}\)is the resultant velocity.

Consider the above equation,

Substitute\(28.57\;m/s\)for\({v_x}\),\( - 34.34\;m/s\)for\({v_y}\)into the above equation,

\(\begin{aligned}{c}v &= \sqrt {{v_x}^2 + {v_y}^2} \\ &= \sqrt {{{\left( {28.57\;m/s} \right)}^2} + {{\left( { - 34.34\;m/s} \right)}^2}} \\ &= 44.7\;m/s\end{aligned}\)

So, the resultant value of the velocity is\(44.7\;m/s\).

Now determine the direction of the resultant velocity,

\({\theta _x} = {\tan ^{ - 1}}\left| {\frac{{{v_y}}}{{{v_x}}}} \right|\)

Substituting the value of horizontal and vertical velocity,

\(\begin{aligned}{c}{\theta _x} &= {\tan ^{ - 1}}\left| {\frac{{ - 34.34}}{{28.57}}} \right|\\ &= 50.2^\circ \end{aligned}\)

Therefore the value of the resultant velocity is \(44.7\;m/s\) and its direction is \(50.2^\circ \).