Kicking a Ball: Solving for Height, Time, & Distance

In summary, the girl kicks a ball straight up with an initial velocity of 3 m/s, making contact with the ball 1 m above the ground. The ball reaches a maximum height of 0.5408 m and takes 1.174 seconds to hit the ground. If the girl had given the ball an initial x-component of velocity of 2 m/s, it would land 3.348 m away from her. The ball would hit the ground with a velocity of 2 m/s.
  • #1
maiamorbific
5
0

Homework Statement


Girl kicks ball straight up, makes contact with ball 1 m above ground, initial velocity = 3 m/s
a.) How high does the ball go?
b.) How long until it hits the ground?
c.) If she had given it an initial x-component of velocity of 2 m/s (same y-component), how far would the ball land from her?
d.) With what velocity would it hit the ground?

Homework Equations


vf2 = vi2 + 2ay(yf-yi)
vf = vi + at
xf = xi + vit + 1/2axt2

The Attempt at a Solution


a.) vf2 = vi2 + 2ay(yf-yi)
0 = 32 + 2(9.8)(yf - 1)
-9 = 19.6yf - 19.6
10.6/19.6 = yf
yf = 0.5408 m
This seems like a really small number; shouldn't it be higher?

b.) vf = vi + at
0 = vi + -gt
t = vi/g
t = 3/9.8 = 0.306 s
0.306 s * 2 = 0.612 s (because the flight curve is symmetric)

I'm not really sure what equations to use for parts c and d, neither am I sure if parts a and b are correct.
Thank you very much!
 
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  • #2
a) check your convention is gravity positive or negative?

b) you need to think about where she initially kicked it from and whether time up == time down for entire trip or just for the symmetric part.

c) What do you need to know about the ball to determine how far it traveled given no air resistance?
hint: it's none of those equations, it's simpler than that.
 
  • #3
Thank you for your help so far.

a.) Changing a to -9.8 gives a final y-position of 2.837 m.
b.) Using the eqn. yf = yi + vit + 1/2at2 makes more sense because it takes into account the non-zero starting position of 1 m.
Therefore, t = 1.174 s
question:
Do I have to multiply this answer by 2? (I don't think so, because doesn't the quadratic equation represent the entire parabola of the flight curve?)

question:
c.) I don't know what you mean by not having to use any eqn.
I used xf = xi + vit + 1/2at2
xf = 1 + 2(1.174) + 0 (since no a in x-direction)
xf = 3.348 m
Is this not correct?

question:
d.) Would vf = vi because there no acceleration in x-direction?
Would the answer be 2 m/s?
 
  • #4
for c I was thinking of even more simply: d = v*t, which is just a simplified version of the one you used.

d) you still need to take y velocity into account here...
 

1. How do you calculate the height of a ball being kicked?

To calculate the height of a ball being kicked, you will need to use the equation h = v2sin2(θ)/2g, where h is the height, v is the initial velocity, θ is the angle of the kick, and g is the acceleration due to gravity. Plug in the known values and solve for h.

2. How do you calculate the time it takes for a ball to reach its peak height?

The time it takes for a ball to reach its peak height can be calculated using the equation t = v*sin(θ)/g, where t is the time, v is the initial velocity, θ is the angle of the kick, and g is the acceleration due to gravity. Plug in the known values and solve for t.

3. How do you determine the distance a ball will travel when kicked?

To determine the distance a ball will travel when kicked, you can use the equation d = v2sin(2θ)/g, where d is the distance, v is the initial velocity, θ is the angle of the kick, and g is the acceleration due to gravity. Plug in the known values and solve for d.

4. How does air resistance affect the trajectory of a kicked ball?

Air resistance can have a significant impact on the trajectory of a kicked ball. It can cause the ball to slow down and deviate from its intended path. Factors such as the shape and texture of the ball, as well as the density and humidity of the air, can affect the amount of air resistance experienced by the ball.

5. Is it possible to calculate the exact path of a kicked ball?

It is possible to calculate the exact path of a kicked ball using mathematical equations, but it may not always be accurate due to factors such as air resistance and external forces. In reality, the exact path of a kicked ball may vary slightly from the calculated trajectory.

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