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Basic Projectile: soccer ball kick 
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#1
Jul708, 03:53 PM

P: 6

1. The problem statement, all variables and given/known data
A soccer ball is kicked with an inital speed of 12.0 m/s in a direction of 25.0 degrees above the horizontal. Find the magnitude and direction of its velocity A) 0.250 s and b) 0.500 s after being kicked. C) Is the ball at its greatest height before or after 0.500 s? 2. Relevant equations a = (V_{f}  V_{i})/t d = ((V_{i} + V_{f}f / 2)t d = V_{i}t + 0.5at[super]2[/super] V_{f} = 2ad + V_{i}[super]2[/super] 3. The attempt at a solution I'm really stuck on this. I've gone with: Vh = cos(25)12.0m/s Vh = 10.88 m/s Vi = sin(25)12.0m/s Vi = 5.07 m/s Now I've heard that in class if your initial vertical velocity in these types of problems has to be equal to the final or finishing vertical velocity... So since I have Vi = 5.07 m/s ; I set Vf = 5.07 m/s ...and worked from there, but I am not sure at all if this was correct. From there I tried to work out the displacement ... Vf = 2ad + V_{i}[super]2[/super] 5.07m/s = 2(9.8m/s2)(d) + 5.07[super]2[/super] d = 1.57 m ... and this is where I KNEW (at least I think I know...) that I'm horribly wrong. A kick with an initial vertical velocity of 5.07 m/s and overall 12.0 m/s ... and my displacement is 1.57 m total... If anybody can help get me pointed in the right direction on this ... I would GREATLY appreciate it. 


#2
Jul708, 04:04 PM

P: 686

[tex] v_x = v_0_x +a_x \cdot t = v_0_x =12 \cos(25^{o}) [/tex]
[tex] v_y= v_0_y + a_y \cdot t =12sin(25^{o})9.81 \cdot t [/tex] [tex]v=\sqrt{v_x^2+v_y^2} [/tex] 


#3
Jul708, 04:50 PM

P: 6




#4
Jul708, 07:00 PM

P: 159

Basic Projectile: soccer ball kick
those equations tell you the final answer... that equation for the v he gave you was the final magnitude of the velocity. The a is not unknown because it is 9.8m/s^2 (acceleration of gravity in Y direction) the T is unknown because you have to plug in the values of t you want to solve for... Vx= Vox*cos(25) Vy= Voy*sin(25)(9.8m/s^2)(t) time doesnt have to be factored into the Vx because that speed is at a constant speed in the x direction until it hits the ground because there are no forces acting on it (assuming no air friction). For Vy does need time factored into it because gravity is acting on it. So plug in the time values for t that you need answers for and use the equation given by dirk to find the magnitude of the velocity... V= sqrt(Vx^2+Vy^2) 


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