Question

A soccer ball is kicked from the ground with an initial speed of${}^{\mathbf{19}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$at an upward angle of${}^{\mathbf{45}\mathbf{°}}$. A player${}^{\mathbf{55}\mathbf{}\mathbf{m}}$ away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground?

Short answer

Average speed if player has to meet the ball just before it hits the ground is$\left(\begin{array}{c}-5.8\mathrm{m}/\mathrm{s}\end{array}\right)\hat{\mathrm{i}}$

Step-by-step solution

Given information

$$\begin{gathered} \stackrel{\rightharpoonup }{v_0}=19.5m/s \\ \theta =45° \end{gathered}$$

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

To understand the concept of projectile motion

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,

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

(i)

The horizontal displacement in projectile motion is given by,

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

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

Where

$∆x=d_x-d_{x0}$

(a) to find the average speed if player has to meet the ball just before it hits the ground

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),

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

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

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