Question

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory and notices that it lands, two seconds after being kicked, about 20 m away to the north. Assume that air resistance is negligible, and plot the horizontal and vertical components of the ball's velocity as a function of time. Consider only the time that the ball is in the air, after being kicked but before landing. Take "north" and "up" as the positive x ‑ and y ‑directions, respectively, and use g≈10g≈10 m/s2 for the downward acceleration due to gravity

Answer 1

Let $u$ be the initial velocity of the soccer ball at an angle of inclination of $\theta_0$ with the positive x-axis.

Given that:

$\theta_0=45^{\circ}$

The horizontal distance covered by the projectile=20 m

Time of flight, $t_f=2$ seconds

Acceleration due to gravity, $g= 10 m/s^2$ downward.

As "north" and "up" as the positive x ‑ and y ‑directions, respectively.

So, $g= -10 m/s^2$

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, $u_x=u\cos\theta_0$.

The horizontal distance covered by the projectile $= u_x\times t_f$

$\Rightarrow u_x\times t_f=20$

$\Rightarrow u_x\times 2=20$

$\Rightarrow u_x=10$ 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,  $u\cos(45^{\circ})=10$ [as $u_x=u\cos\theta_0$]

$\Rightarrow u=10\sqrt{2}$ m/s.

The vertical component of the initial velocity,

$u_y= u\sin\theta_0$

$\Rightarrow u_y=10\sqrt{2}\sin(45^{\circ})$

$\Rightarrow u_y=10$ m/s

Let v be the vertical component of the velocity at any time instant t.

From the equation of motion,

$v=u+at$

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 $u=u_y, a= -10 m/s^2$.

So at any time instant, t.

$v=u_y+(-10)t$

$\Rightarrow v=10-10t$

The vertical component of the velocity, v, is the function of time and related as $v=10-10t$.

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).