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Q: Ball kicked into the air at an angle of 34.0 degrees above the horizontal the initial velocity of the ball is 25.0 ms How long is the soccer ball in the air?

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Initial velocity is 10 m/s in the direction it was kicked. Final velocity is 0, when friction and air resistance finally causes it to come to a halt.

You kicked the rock with an initial velocity of 3.4 m/s.

If it is kicked at 30 degrees to the horizontal then it will travel 27.7 metres.

initial velocity of the kick = 28.06 m/s

When a ball is kicked at an angle, there is no acceleration along the horizontal direction (since there isn't any force along the direction ,ignoring viscous forces), so , its velocity along the horizontal direction remains unchanged.... according to the 1st law , velocity changes only when a net resultant force is applied on the ball , so , Newton's law is valid. only the initial angle of kick and the vertical component of velocity are mainly responsible for the distance travelled by the ball horizontally....

At the top of it path or anywhere else the Earth's gravity accelerates the ball downward. Regardless what the horizontal mothion is (disregarding air resistance), there are no accelerations after the kick. A ball kicked straight up to the same height will hit the ground at the same time as a ball kicked at 45 degrees.

Velocity = Distance/Time = 28/1.4 = 20 m/s

When it's at its maximum height its speed will be zero.

The record speed for a soccer ball to have been kicked by a pro soccer player is 110 mph.

I assume the question is referring to a ball say that is being kicked, in this case some fairly simplistic physics determines that 45 degrees is the perfect angle, this calculation relies upon the fact that the ground is at the same height when it was kicked as when to when it comes to rest. (Delta Yx =0. Hence a level ground) However... If you are talking about a shot put for instance there is a different story, the optimum angle for this type of throw would be around 42 degrees. This value is different from the first due to the differential in heights between Y(x0) and Y(xfinal) where X is distance, Y states the vertical plane and 0 when t=0 (initial height) and then final when Vx&Vy=0 (aka rest). A key point to remember in the proof of this is that velocity, time and distance are all independent of mass. Regardless of how you choose to calculate the initial velocity the explanation stated above is always true for the angle both in a vacuum and in a real world environment. You didn't ask for a scientific proof so in summary: 45 degrees for a ground object (eg. football/soccer) 42 degrees for an object thrown from torso height (eg. shotput) - Approximation

It never happened. Einstein left his Munich school in order to study in Zurich, where he flunked his initial entrance exams.

I/We/You/They have kicked He/She/It has kicked

No, the word kicked is not an adverb.The word kicked is a verb.

you cant find an average, because NO-ONE kicks at exactly the same amount of speed and force as any other person.

In soccer, the soccer ball has projectile motion when it is kicked, head-butted, when the speed changes, and when the velocity changes.

She is almost kicked out but Logan and the boys helped her not get kicked out.

The British kicked out the acadiens

I/We/You/They have kicked He/She/It has kicked

This is a ballistic motion problem. A 30-kg ball -- man, that's one heavy damn ball -- is kicked at an angle of 45 degrees to the horizontal and travels a distance of 20 meters. Well, aside from a broken toe, what can we determine? If I recall from trajectory/ballistics problems, the range of a projectile is given by the following formula: R = 2VxVy/g, where Vx and Vy are the horizontal and vertical components of the velocity, respectively. Vx = VcosA and Vy = VsinA, where A is the take-off angle, 45 degrees in this case. That is very convenient, because both components are, therefore, equal in magnitude: Vx = Vy = V/SQRT(2). So, making the substitutions into the equation above, we get R = V2/g. Solving for V, we get V = SQRT(Rg). Substituting, we get V = SQRT(196) = 14 (m/s). Note how the mass of the ball had no bearing on the answer. I decided to fiddle a bit more with the problem and use only the basic formulas used in high-school physics that deal with displacement (distance), velocity, and acceleration. The main formula for calculating displacement, d, is: d(t) = do + Vot + (1/2)at2, where do is the initial displacement, Vo is the initial velocity, and a is acceleration. And there is one other basic formula that will come in handy later: V(t) = Vo + at. In English, that means the velocity at any time, t, can be found by multiplying the acceleration by t and adding it to the initial velocity. For this problem, there are two main equations, one for displacement in the horizontal direction, dx, and one for displacement in the vertical direction, dy. For both we assume that the initial displacement, do, is zero, since the initial point of flight -- the kick-off point -- is the origin and represents zero displacement in either direction (x or y). So, dox = doy = 0. We also know there is no acceleration in the xdirection, and the acceleration in the y direction is the acceleration due to gravity, g, which is negative, that is, directed downward. So, dx(t) = dox + Voxt + (1/2)at2 = 0 + Voxt + 0 = Voxtanddy(t) = doy + Voyt + (1/2)at2 = 0 + Voyt - (1/2)gt2 = Voyt - (1/2)gt2So, we have the following: (Equ. 1) dx(t) = Voxtand (Equ. 2) dy(t) = Voyt - (1/2)gt2But we must now get our brains around a few observations and facts of the problem. Since the take-off angle is 45 degrees to the horizontal, we know that the horizontal and vertical components of the initial velocity are equal in magnitude. (Their directions, of course, are perpendicular to each other.) So, we can write Vox = Voy. This fact will come in handy later. We also know that at some later time, T, the ball will strike the ground 20 meters away. So, substituting for d and t in Equ. 1 above, we have 20 = VoxT. Solving for T, we write: (Equ. 3) T = 20/Vox Further, we also know that the ball reaches the highest point in its trajectory half way through its trip, at t = T/2. And we also know that at its highest point, the vertical velocity is zero. Recall the earlier formula: V(t) = Vo + at. That will be useful now. Using that formula, we can write the equation for the velocity in the vertical direction like this: Vy(t) = Voy - gt And since we already established that the vertical velocity is zero at t = T/2, the equation becomes: Vy(T/2) = Voy - g * (T/2) = 0 Solving for T, we get: (Equ. 4) T = 2Voy/g But now we have two independent equations (Equ. 3 and Equ. 4) for T, which is very cool. We can now write: 20/Vox = 2Voy/g Rearranging the terms, we can write: VoxVoy = 10g But since Vox = Voy, we can write V2 ox = 10g, or Vox = SQRT(10g) = SQRT(98) = 9.9. Uh-oh, that's not 14, like we calculated earlier. Nope, it isn't. That's because we calculated only the x component of the velocity. The magnitude of the velocity of the ball is the vector sum of the x and ycomponents. Since they are equal and at right angles to each other, we can easily use the Pythagorean theorem to calculate it: V2 = V2ox + V2oy V = SQRT(V2ox + V2oy) V = SQRT(10g + 10g) = SQRT(196) = 14 m/s. Bingo!

No he did not get kicked out of baseball.

No you cannot get kicked out a state. No matter what you do.

she was kicked out

The past tense is kicked.

2.08 seconds

Yes, that is the correct spelling of the word kicked.Some example sentences are:He kicked the ball over the fence.I was kicked out of the bar.I will tell mum you kicked me.