Question

Short answer

(d) The velocity (including both the horizontal and vertical components) of the ball just before it hits the ground is \(44.7\;m/s\). The velocity is directed \(50.2^\circ \) down the x-axis.

Step-by-step solution

Identification of the given data

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

Understanding the concept

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.

Step 3: Determine the resultant and its direction:

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 \(50.2^\circ \).