Question

In a soccer game a midfielder kicks the ball from one touchline directly toward the other with a velocity of 14.0 m/s (along the positive x-axis). After the ball travels for 2.50 s, a striker makes an acrobatic kick that gives the ball a new velocity in the y direction, and after 0.900 s the ball passes the diving goalkeeper and goes into the net. If the total displacement of the ball is 40.2 m at 29.5° above the x positive axis, with what speed did the striker kick the ball?

Answer 1

Displacement components.

The vertical displacement is :

$sin(29.5^{0})=\frac{Y_{T}}{r_{T}}\\ Y_{T}=r_{t}sin(29.5^{0}) \\Y_{T}=40.2m*sin(29.5^{0})=19.8m$

The horizontal displacement is

$cos(29.5^{0})=\frac{X_{T}}{r_{T}}\\ X_{T}=r_{t}cos(29.5^{0}) \\X_{T}=40.2m*cos(29.5^{0})=35.0m$

The horizontal displacement of the first 2.50 seconds is:

$X_{1}=V_{ox}*t\\ X_{1}=14\frac{m}{s}*2.5s=35.0m$

The displacement in the other 0.900s is

$X_{2}=X_{T}-X_{1}\\X_{2}=40.2m-35.0m=5.2m$

The horizontal velocity printed by the striker is

$V_{x}=\frac{X_{2}}{t}\\V_{x}=\frac{5.2m}{0.900s}=5.78\frac{m}{s}$

The horizontal movement is independent of the vertical movement.

The vertical position is given by:

$Y=Y_{o}+V_{oy}*t+\frac{1}{2}*a_{y}*t^{2}$

Where

• Initial position $Y_{o}=0m$
• Final position $Y=19.8m$
• Vertical acceleration a$a_{y}=-9.8\frac{m}{s^{2}}$
• The vertical velocity $V_{y}=??$
• Time. $t=0.900s$

Expression of the vertical velocity

$Y=V_{oy}*t+\frac{1}{2}*a_{y}*t^{2}\\V_{oy}=\frac{Y-\frac{1}{2}*a_{y}*t^{2}}{t} \\$

Numerically evaluating

$V_{oy}=\frac{19.8m-\frac{1}{2}*-9.8\frac{m}{s^{2}}*(0.900s)^{2}}{0.900s}=17.6\frac{m}{s}$

The vertical velocity given by the striker is $V_{oy}=17.6\frac{m}{s}$

The magnitude of the velocity given by the striker is:

$V^{2}=V_{ox}^{2}+V_{oy}^{2}\\ V=\sqrt{(5.78\frac{m}{s})^{2}+(17.6\frac{m}{s})^{2}}\\ V=18.5\frac{m}{s}$

Finally

$\boxed{V=18.5\frac{m}{s}}$