What is the Range of a Soccer Player's Kick at 20m/s and 15° Angle?

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A soccer player kicks a stationary ball at 20 m/s at a 15° angle, prompting a discussion on calculating the range. The initial vertical speed is determined to be 5.18 m/s, leading to a time to reach maximum height of 0.53 seconds. The horizontal speed, calculated as 19.32 m/s, remains constant due to the absence of horizontal forces. The range can be found using the formula R = Vo² sin(2θ) / g, with Vo being the initial speed of 20 m/s. The calculations confirm that the approach to finding the range is correct, emphasizing the importance of both vertical and horizontal components in projectile motion.
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Homework Statement


a soccer player kicks a stationary ball at a speed of 20m/s at an angle of 15\circ to the horizontal.


Homework Equations



i think it would be Rmax=Vo2/g ?

The Attempt at a Solution


19.32 2/9.8=38.09m ??

I'm really lost can someone help me please!
 
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From the angle work out the vertical speed
Then think about how the vertical speed changes with time

hint. what happens at the top of the flight?
 
mmm... so i calculate the vertical speed: 20sin15=5.18m/s? and with that i can find at what time it reaches the max height by doing V=Vo+at: V=5,18/9.8= 0.53s do i have to multiply the time by 2 and then do the same thing to find the horizontal range?

X=Xo+Vot-1/2gt(square) where g=0 and Xo=0 too, so it would be X=19.32*(2*0.53s)?
 
Whats the acceleration in the horizontal direction?

hint. if you ignore air resistance what forces act horizontally
 
No idea...im sooo lost! i looked in my book and found this formula R=Vo(square)sin2teta/g omg I am freaking out! what Vo i have to take?!? 20m/s? the Vy? or the Vx?
 
is it (19.32m/s)^2 sin(30)/9.8m/s^2= 19.04m?
 
hmm...the only force i see is the Vx=20m/s cos 15=19.32m/s it's the only one horizontaly
 
yellow_angel said:
mmm... so i calculate the vertical speed: 20sin15=5.18m/s? and with that i can find at what time it reaches the max height by doing V=Vo+at: V=5,18/9.8= 0.53s do i have to multiply the time by 2 and then do the same thing to find the horizontal range?

X=Xo+Vot-1/2gt(square) where g=0 and Xo=0 too, so it would be X=19.32*(2*0.53s)?

All of this is correct!
(Although, you didn't show where or how you got 19.32).
 
yellow_angel said:
hmm...the only force i see is the Vx=20m/s cos 15=19.32m/s it's the only one horizontaly
Correct, there is no horizontal force, so no aceleration, so speed is constant
 
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