Explanation:
It is given that,
Initial velocity of the soccer player, u = 30 m/s
It is launched at an angle of 45 degrees with the horizontal.
1. Time of flight :
The time taken by the projectile in the air and to comes to ground is called the time of flight. It is given by :
[tex]T=\dfrac{2u\ sin\theta}{g}[/tex]
[tex]T=\dfrac{2(30)\ sin(45)}{9.8}[/tex]
T = 4.32 seconds
2. Range of the projectile
The horizontal distance covered by the projectile is called its range. It is given by :
[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]
[tex]R=\dfrac{(30)^2\ sin2(45)}{9.8}[/tex]
R = 91.83 meters
3. Maximum height reached by the ball is given by :
[tex]H=\dfrac{(u\ sin\theta)^2}{2g}[/tex]
[tex]H=\dfrac{(30\times \ sin(45))^2}{2\times 9.8}[/tex]
H = 22.95 meters
Hence, this is the required solution.
