Respuesta :
a) v.h=v*cos(63)=7.26 m/s
b) v.v=v*sin(63)=14.26 m/s
c) t=v.v/g=14.26/9.8=1.46 s since time the body takes to go upward is defined by the acceleration due to gravity and the vertical component of the speed
d) h=1/2*gt^2=7.15 m
e) T=2t=2.92 s
f) R=v.h*t=7.26*1.46=10.6 m since horizontal displacement is defined by the horizontal component of the speed only.
a) The horizontal component of the launch velocity is approximately 6.810 meters per second.
b) The vertical component of the launch velocity is approximately 13.365 meters per second.
c) The soccer ball does take 1.363 seconds to reach its highest point.
d) The soccer ball reaches a maximum height of 9.107 meters.
e) The soccer ball has a total amount of time of 2.725 seconds in the aire.
f) The soccer ball is 18.557 meters far from the soccer player.
a) The soccer ball experiments a Parabolic Motion, which is a combination of horizontal motion at uniform Velocity and vertical Free-Fall motion. The horizontal component of the launch Velocity ([tex]v_{o,x}[/tex]), in meters per second, is found by this formula:
[tex]v_{o, x} = v_{o}\cdot \cos \alpha[/tex] (1)
Where:
[tex]v_{o, x}[/tex] - Initial velocity of the soccer ball, in meters per second.
[tex]\alpha[/tex] - Launch angle, in sexagesimal degrees.
If we know that [tex]v_{o} = 16\,\frac{m}{s}[/tex] and [tex]\alpha = 63^{\circ}[/tex], then the horizontal component of the velocity is:
[tex]v_{o,x} = \left(16\,\frac{m}{s} \right)\cdot \cos 63^{\circ}[/tex]
[tex]v_{o,x} \approx 6.810\,\frac{m}{s}[/tex]
The horizontal component of the launch velocity is approximately 6.810 meters per second.
b) And the vertical component of the launch Velocity ([tex]v_{o,y}[/tex]), in meters per second, in found by this formula:
[tex]v_{o,y} = v_{o}\cdot \sin \alpha[/tex] (2)
If we know that [tex]v_{o} = 16\,\frac{m}{s}[/tex] and [tex]\alpha = 63^{\circ}[/tex], then the horizontal component of the velocity is:
[tex]v_{o,y} = \left(16\,\frac{m}{s} \right)\cdot \sin 63^{\circ}[/tex]
[tex]v_{o,y} \approx 13.365\,\frac{m}{s}[/tex]
The vertical component of the launch velocity is approximately 13.365 meters per second.
c) The soccer ball reaches the maximum Height ([tex]y[/tex]), in meters, when vertical Velocity ([tex]v_{o,y}[/tex]), in meters per second, is zero, the Kinematic formula needed is described below:
[tex]v_{y} = v_{o,y}+g\cdot t[/tex] (3)
Where:
[tex]g[/tex] - Free-fall acceleration, in meters per square second.
[tex]t[/tex] - Time, in seconds.
If we know that [tex]v_{o,y} \approx 13.365\,\frac{m}{s}[/tex], [tex]v_{y} = 0\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the time taken by the soccer ball before reaching its highest point:
[tex]13.365\,\frac{m}{s} + \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t = 0\,\frac{m}{s}[/tex]
[tex]t = 1.363\,s[/tex]
The soccer ball does take 1.363 seconds to reach its highest point.
d) The maximum Height of the soccer ball can be found by using the following Kinematic expression:
[tex]v_{y}^{2} = v_{o,y}^{2}+2\cdot g\cdot y[/tex] (4)
If we know that [tex]v_{o,y} \approx 13.365\,\frac{m}{s}[/tex], [tex]v_{y} = 0\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the maximum height reached by the soccer ball is:
[tex]\left(0\,\frac{m}{s} \right)^{2} = \left(13.365\,\frac{m}{s} \right)^{2} + 2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot y[/tex]
[tex]y = 9.107\,m[/tex]
The soccer ball reaches a maximum height of 9.107 meters.
e) The total amount of Time that the soccer ball is in the air can be found from this expression and based on the fact that final Height is zero:
[tex]y = v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}[/tex] (5)
If we know that [tex]v_{o,y} \approx 13.365\,\frac{m}{s}[/tex], [tex]y = 0\,m[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then the total amount of Time taken by the soccer ball is:
[tex]\left(13.365\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2} = 0\,m[/tex]
[tex](13.365-4.904\cdot t)\cdot t = 0[/tex]
There are two solutions:
[tex]t = 0\,s\,\lor\,t = 2.725\,s[/tex]
The second root represents the total amount of time taken by the soccer ball.
The soccer ball has a total amount of time of 2.725 seconds in the aire.
f) The horizontal Distance travelled by the soccer ball ([tex]x[/tex]), in seconds, is calculated from the following Kinematic expression:
[tex]x = v_{o,x}\cdot t[/tex] (6)
If we know that [tex]v_{o,x} \approx 6.810\,\frac{m}{s}[/tex] and [tex]t = 2.725\,s[/tex], then the horizontal distance travelled by the soccer ball is:
[tex]x = \left(6.810\,\frac{m}{s} \right)\cdot (2.725\,s)[/tex]
[tex]x = 18.557\,m[/tex]
The soccer ball is 18.557 meters far from the soccer player.
Please see this question for further details on: https://brainly.com/question/16992646
