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* t_f`

`\Rightarrow u_x* t_f=20`

`\Rightarrow u_x* 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√(2)` m/s.

The vertical component of the initial velocity,

`u_y= u\sin\theta_0`

`\Rightarrow u_y=10√(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).