91Ó°ÊÓ

$$\begin{gathered} \stackrel{â‡\P Ä}{v_0}=19.5m/s \\ \mathrm{θ}=45° \end{gathered}$$

Initial displacement in horizontal direction${}^{d_{x0}=55m}$

This problem deals with the projectile motion of an object. The projectile motion of an object is the motion of an object thrown into the air, subject to the acceleration due to gravity. Using the standard formula for the range of the projectile the horizontal distance can be computed.

Further, using the formula for the second kinematic equation, time of flight of a ball can be found. Finally, using the formula for average velocity, average speed of the soccer ball can be calculated.

Formulae:

The vertical displacement in projectile motion is given by,

$d_x=\left(v_0\sin \mathrm{θ}\right)+\frac{1}{2}a{t}^2$

(i)

The horizontal displacement in projectile motion is given by,

${}^{d_x=\left(v_0\cos \mathrm{θ}\right)t}$(ii)

${\stackrel{→}{v}}_{avg}=\frac{∆x}{∆t}$ (iii)

Where

$∆x=d_x-d_{x0}$

Using equation (i) the time will be,

$t^2=\frac{\left[\left(19.5m/s\right)\sin \left(45\right)\right]2}{9.8m/s^2}$

Thus

$t=2.81s$

Now, for this time, the horizontal distance is given by equation (ii),

$$\begin{gathered} d_x=\left[\left(19.5m/s\right)\cos \left(45\right)\right]×2.01 \\ {d}_x=38.7m \end{gathered}$$

Therefore, using equation (iii) the average velocity of the player is,

$$\begin{gathered} {\stackrel{→}{\mathrm{v}}}_{\mathrm{avg}}=\frac{\left(\begin{array}{c}38.7\mathrm{m}\end{array}\right)\hat{\mathrm{i}}-\left(\begin{array}{c}55\mathrm{m}\end{array}\right)\hat{\mathrm{j}}}{2.81\mathrm{s}} \\ {\stackrel{→}{\mathrm{v}}}_{\mathrm{avg}}=\left(\begin{array}{c}-5.8\mathrm{m}/\mathrm{s}\end{array}\right)\hat{\mathrm{i}} \end{gathered}$$