Physics, 15.10.2020 14:01, reneek272

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

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answered: hrijaymadathil

Let 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).

Emmy kicks a soccer ball up at an angle of 45° over a level field. She watches the ball's trajectory

answered: Guest

hi there if you are asking for conformation that wavelingth is the correct answer is wrong it is the answer is frequency. : )

i hope this you out, have a wonderful day

answered: Guest

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this. hope that you like the video and that it explains more in detail to you.

that is the link.