Let [tex]u[/tex] be the initial velocity of the soccer ball at an angle of inclination of [tex]\theta_0[/tex] with the positive x-axis.
Given that:
[tex]\theta_0=45^{\circ}[/tex]
The horizontal distance covered by the projectile=20 m
Time of flight, [tex]t_f=2[/tex] seconds
Acceleration due to gravity, [tex]g= 10 m/s^2[/tex] downward.
As "north" and "up" as the positive x ‑ and y ‑directions, respectively.
So, [tex]g= -10 m/s^2[/tex]
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, [tex]u_x=u\cos\theta_0[/tex].
The horizontal distance covered by the projectile [tex]= u_x\times t_f[/tex]
[tex]\Rightarrow u_x\times t_f=20[/tex]
[tex]\Rightarrow u_x\times 2=20[/tex]
[tex]\Rightarrow u_x=10[/tex] 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, [tex]u\cos(45^{\circ})=10[/tex] [as [tex]u_x=u\cos\theta_0[/tex]]
[tex]\Rightarrow u=10\sqrt{2}[/tex] m/s.
The vertical component of the initial velocity,
[tex]u_y= u\sin\theta_0[/tex]
[tex]\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})[/tex]
[tex]\Rightarrow u_y=10[/tex] m/s
Let v be the vertical component of the velocity at any time instant t.
From the equation of motion,
[tex]v=u+at[/tex]
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 [tex]u=u_y, a= -10 m/s^2[/tex].
So at any time instant, t.
[tex]v=u_y+(-10)t[/tex]
[tex]\Rightarrow v=10-10t[/tex]
The vertical component of the velocity, v, is the function of time and related as [tex]v=10-10t[/tex].
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).
