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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 = Voxt

and

dy(t) = doy + Voyt + (1/2)at2 = 0 + Voyt - (1/2)gt2 = Voyt - (1/2)gt2

So, we have the following: (Equ. 1) dx(t) = Voxt

and (Equ. 2) dy(t) = Voyt - (1/2)gt2

But 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!

Q: What is the take-off velocity of a 30-kg ball that is kicked at an angle of 45 degrees to the horizontal and travels 20 meters?

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