Let
be the initial velocity of the soccer ball at an angle of inclination of
with the positive x-axis.
Given that:
![\theta_0=45^{\circ}](https://tex.z-dn.net/?f=%5Ctheta_0%3D45%5E%7B%5Ccirc%7D)
The horizontal distance covered by the projectile=20 m
Time of flight,
seconds
Acceleration due to gravity,
downward.
As "north" and "up" as the positive x ‑ and y ‑directions, respectively.
So, ![g= -10 m/s^2](https://tex.z-dn.net/?f=g%3D%20-10%20m%2Fs%5E2)
As the acceleration due to gravity is in the vertical direction, so the horizontal component of the initial velocity remains unchanged.
The x-component of the initial velocity,
.
The horizontal distance covered by the projectile ![= u_x\times t_f](https://tex.z-dn.net/?f=%3D%20u_x%5Ctimes%20t_f)
![\Rightarrow u_x\times t_f=20](https://tex.z-dn.net/?f=%5CRightarrow%20u_x%5Ctimes%20t_f%3D20)
![\Rightarrow u_x\times 2=20](https://tex.z-dn.net/?f=%5CRightarrow%20u_x%5Ctimes%202%3D20)
m/s
So, the horizontal component of the velocity is 10 m/s which is constant and the graph has been shown in the figure (i).
Now,
[as
]
m/s.
The vertical component of the initial velocity,
![u_y= u\sin\theta_0](https://tex.z-dn.net/?f=u_y%3D%20u%5Csin%5Ctheta_0)
![\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})](https://tex.z-dn.net/?f=%5CRightarrow%20u_y%3D10%5Csqrt%7B2%7D%5Csin%2845%5E%7B%5Ccirc%7D%29)
m/s
Let v be the vertical component of the velocity at any time instant t.
From the equation of motion,
![v=u+at](https://tex.z-dn.net/?f=v%3Du%2Bat)
where u: initial velocity, v: final velocity, a: constant acceleration, and t: time taken to change the velocity from u to v.
In this case, we have
.
So at any time instant, t.
![v=u_y+(-10)t](https://tex.z-dn.net/?f=v%3Du_y%2B%28-10%29t)
![\Rightarrow v=10-10t](https://tex.z-dn.net/?f=%5CRightarrow%20v%3D10-10t)
The vertical component of the velocity, v, is the function of time and related as
.
This is a linear equation.
At 2 second, the vertical component of the velocity
v=10-10x2=-10 m/s.
The graph has been shown in figure (ii).