Football Projectile Calculations

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In summary, the football reaches its maximum height of 16.53 meters after 1.1836 seconds. It then takes 3.673 seconds to return to its original level, and travels a horizontal distance of 91.83 meters during this time.
  • #1
courtrigrad
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A football player throws a football with an initial upward velocity component of 18.0 m/s and a horizontal velocity component of 25.0 m/s. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after being thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far as it traveled horizontally during this time?

(a) Ok so I know that [itex] v_{x}_{0} = 25.0 m/s, v_{y}_{0} = 18.0 m/s [/itex]. So for part (a) we know that the vertical component of velocity [itex] v_{y} = 0 [/itex]. So would I use [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} [/itex] where [itex] v_{y}_{0} = v_{0} = 18.0 m/s, v_{y} = 0, \theta = 90, g = 9.8 m/s^{2} [/itex]?

(b) Would I use [itex] y = (v_{0}\sin\theta_{0})t - \frac{1}{2}gt^{2} [/itex] and plug in the time?
(c) wouldn't the times for part(a) and part (c) be equal neglecting air resistance?
(d) I would just find the time for the football to travel the horizontal distance and use the equation [itex] R = (v_{0}\cos\theta_{0})t_{2} [/itex]?

Thanks
 
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  • #2
plugpoint said:
A football player throws a football with an initial upward velocity component of 18.0 m/s and a horizontal velocity component of 25.0 m/s. (a) How much time is required for the football to reach the highest point of the trajectory? (b) How high is this point? (c) How much time (after being thrown) is required for the football to return to its original level? How does this compare with the time calculated in part (a)? (d) How far as it traveled horizontally during this time?

(a) Ok so I know that [itex] v_{x}_{0} = 25.0 m/s, v_{y}_{0} = 18.0 m/s [/itex]. So for part (a) we know that the vertical component of velocity [itex] v_{y} = 0 [/itex]. So would I use [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} [/itex] where [itex] v_{y}_{0} = v_{0} = 18.0 m/s, v_{y} = 0, \theta = 0, g = 9.8 m/s^{2} [/itex]?

(b) Would I use [itex] y = (v_{0}\sin\theta_{0})t - \frac{1}{2}gt^{2} [/itex] and plug in the time?
(c) wouldn't the times for part(a) and part (c) be equal neglecting air resistance?
(d) I would just find the time for the football to travel the horizontal distance and use the equation [itex] R = (v_{0}\cos\theta_{0})t_{2} [/itex]?

Thanks
Remember that v=<25.0, 18.0>m/s. You can either plug that in and find your vertex (similar to what you suggested above) or differentiate and set equal to zero (if you don't know how to take a derivative yet, forget I said that). To find the time it takes for it to hit the ground, you set the y equation equal to zero (do you see why?) You will have two roots to this equation, one being your initial time and the other being the time you need to find. I hope this helps.

Alex
 
  • #3
yeah I think I got it. Is this correct:

(a) [itex] t = \frac{v_{0}\sin\theta_{0} - v_{y}}{g} = \frac{18.0 m/s}{9.8 m/s^{2}}, t = 1.1836 s [/itex]

(b) [itex] y_{max} = (18 m/s)(1.86) - \frac{1}{2}(9.8 m/s^{2})(1.836)^{2} = 16.53 m [/itex]
(c) [itex] t_{2} = \frac{2v_{0}\sin\theta_{0}}{g} = \frac{ 36 m/s}{9.8 m/s^{2}} = 3.673 s [/itex].
(d) [itex] R = (v_{0}\cos\theta_{0})t_{2} = (25 m/s)(3.673 s) = 91.83 m [/itex]

Thanks
 
  • #4
is this correct?
 
  • #5
You upward velocity is 18 m/s, decresaing by 9.8 m/s², so time to reach zero velocity, at which the foorbal is at its maximum height is,

t = Vy/g
t = 18/9.8
t = 1.837 s
========

(b) is correct

(c) is correct

(d) is correct
 
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1. What is the "Football Projectile Problem"?

The "Football Projectile Problem" is a physics problem that involves calculating the trajectory of a football as it is kicked or thrown through the air. It takes into account factors such as the initial velocity, angle of projection, and air resistance to determine the path of the football.

2. Why is the "Football Projectile Problem" important?

The "Football Projectile Problem" is important because it helps us understand the physics behind the movement of a football, which can then be applied to other sports and real-life situations. It also allows us to analyze and improve techniques for throwing or kicking a football, such as determining the optimal angle for maximum distance.

3. What equations are used to solve the "Football Projectile Problem"?

The "Football Projectile Problem" can be solved using several equations, including the basic kinematic equations of motion, the equation for projectile motion, and the drag equation to account for air resistance. These equations can be manipulated and combined to determine the final trajectory of the football.

4. How does air resistance affect a football's trajectory?

Air resistance, also known as drag, can significantly affect a football's trajectory. As the football moves through the air, it creates an area of high pressure in front of it and an area of low pressure behind it. This pressure difference creates a force that acts in the opposite direction of the football's motion, causing it to slow down and deviate from its original path.

5. Can the "Football Projectile Problem" be applied to other objects besides footballs?

Yes, the principles and equations used to solve the "Football Projectile Problem" can be applied to any object that is thrown or kicked through the air, such as baseballs, golf balls, or even projectiles launched from a cannon. However, the specific parameters and factors may vary depending on the object's shape, weight, and air resistance.

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