Answer:
[tex]v=14.5(m/s)[/tex].
Explanation:
This is a projectile motion problem, so for solving it, we will require some equations of uniform motion and free-fall.
First, we need to recognize the data that we have so we can start solving the problem.
For Free -Fall:
[tex]g= -9.8(m/s^{2})[/tex]
[tex]\Delta y=2.4(m)[/tex]
For uniform motion:
[tex]\Delta x=17(m)[/tex]
The time that takes to the ball to travel 17m horizontally and to hit the crossbar at 2.4m of height is the same, so time is a common variable for free-fall and uniform motion.
Using the equations:
[tex]\Delta y=v_{y0}t+0.5gt^{2}[/tex] and
[tex]\Delta x=v_{x}t[/tex]
and noticing that
[tex]v_{y0}=v*sin(38)[/tex] and
[tex]v_{x}=v*cos(38)[/tex],
we obtain
[tex]\Delta y=v*sin(38)t+0.5gt^{2}[/tex] and
[tex]\Delta x=v*cos(38)t[/tex]
wich is a system of two equations and to variables that can be easily solve.
Making
[tex]v=\frac{\Delta x}{t*cos(38)}[/tex]
we get
[tex]\Delta y=(\frac{\Delta x}{t*cos(38)})sin(38)t+0.5gt^{2}[/tex],
[tex]\Delta y=\Delta x*tan(38)+0.5gt^{2}[/tex],
[tex]\Delta y-\Delta x*tan(38)=0.5gt^{2}[/tex],
[tex]t=\sqrt{\frac{\Delta y-\Delta x*tan(38)}{0.5g}}[/tex],
[tex][tex]t=\sqrt{\frac{2.4-17*tan(38)}{0.5*(-9.8)}}[/tex]
so
[tex]t=1.50s[/tex],
now the speed can be easily compute from one of our equations. Using
[tex]v=\frac{\Delta x}{t*cos(38)}[/tex],
[tex]v=\frac{17}{1.5*cos(38)}[/tex],
[tex]v=14.5(m/s)[/tex].
The speed of the ball when it left his foot is 11.13 m/s.
The given parameters;
The speed of the projectile will be calculated by applying the formula for maximum height reached by a projectile;
[tex]H = \frac{u^2 sin^2(\theta)}{2g} \\\\u^2 = \frac{2gH}{sin^2(\theta)} \\\\u^2 = \frac{2\times 9.8 \times 2.4}{sin^2 (38)} \\\\u^2 = 123.97 \\\\u = \sqrt{123.97} \\\\u = 11.13 \ m/s[/tex]
Thus, the speed of the ball when it left his foot is 11.13 m/s.
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