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2 years ago

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A player kicks a football (from the ground) at an initial angle of 30°. The football is in the air for 2.4 s before it hits an o

pposing player. The opposing player was a horizontal distance x = 50 m away from where the football was kicked. What was the initial horizontal velocity component of the football?

1 answer:

wolverine [178]2 years ago

8 0

Answer:

The initial horizontal velocity was 21 m/s

Explanation:

Please, see the figure for a better understanding of the problem.

The equation for the position of an object moving in a parabolic trajectory is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g ·t²)

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity

Notice that at time t = 2.4 s the vector "r" is the one in the figure. We know that the x-component of that vector is 50 m. Then using the equation for the x-component of the vector "r", we can calculate the initial velocity:

x = x0 + v0 · t · cos α

Let´s place the center of the frame of reference at the point of the kick so that x0 = 0.

x = v0 · t · cos α

x/t = v0 · cos α

Notice in the figure that v0 · cos 30° = v0x which is the initial horizontal velocity. Remember trigonometry of right triangles:

cos α = adjacent / hypotenuse = v0x / v0

Then:

50 m/ 2.4 s = v0 · cos 30° = v0x

v0x = 21 m/s

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