Did the ball travel far enough to make the goal?

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Homework Help Overview

The problem involves a kicker attempting to determine if a kicked ball travels far enough to clear a goalpost that is 3 meters high, standing 50 meters away. The ball is kicked with an initial velocity of 24.5 m/s at an angle of 35.28 degrees.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations of horizontal and vertical components of the ball's velocity and the time it takes to reach the goalpost. There is also exploration of the height of the ball when it reaches the goalpost and whether it meets the height requirement.

Discussion Status

Some participants express confusion about the phrasing of the question and whether it is asking about height or distance. There are differing interpretations regarding the goal's requirements and the assumptions made about gravity. Guidance is offered to clarify the height of the ball at the goalpost and its relation to the goal height.

Contextual Notes

There are assumptions about the acceleration due to gravity being -10 m/s², and participants note the lack of clarity in the question regarding the conditions for scoring a goal.

azukibean
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1. Problem
The kicker stands 50 m from the goalpost that is 3 m high. She kicks with an initial velocity of 24.5 m/s. She just makes the goal height-wise, but did the ball travel far enough? She kicked at an angle of 35.28 degrees.

2. Homework Equations
S = (1/2)at^2 + V0t + S0
S =V0xt (when V0x is constant)


3. What I Did
24.5cos(35.28) = V0x = 20
24.5sin(35.28) = V0y = 14.15
-5t^2 + 14.15t + 0 = 3
t = 2.6
20*2.6 = 52 m
2 meters to spare

The answer is 1.12 m to spare. Any help would be appreciated! Thanks.
 
Last edited:
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azukibean said:
1. Problem
The kicker stands 50 m from the goalpost that is 3 m high. She kicks with an initial velocity of 24.5 m/s. She just makes the goal height-wise, but did the ball travel far enough? She kicked at an angle of 35.28 degrees.

2. Homework Equations
S = (1/2)at^2 + V0t + S0
S =V0xt (when V0x is constant)


3. What I Did
24.5cos(35.28) = V0x = 20
24.5sin(35.28) = V0y = 14.15
-5t^2 + 14.15t + 0 = 3
t = 2.6
20*2.6 = 52 m
2 meters to spare

The answer is 1.12 m to spare. Any help would be appreciated! Thanks.

What a strange question. Surely if the ball "just made the goal height-wise" the ball is still in the air when it passes over the goal post, the goal has been made, and by definition the ball has traveled far enough.

From the given answer it looks like they meant to ask you by what margin the ball cleared the goal post, and g = 10m/s2 is assumed for the acceleration due to gravity.

Have you stated the question exactly as given to you?
 
a) gravity is assumed to be -10 m/s
b) Kyle will attempt a goal from 50 m in front of the goal post. Kyle boots the ball with an initial velocity of 24.5 m/s at an angle of 35.28. It is right on target but is it long enough? How many meters to spare?
 
Last edited:
azukibean said:
a) gravity is assumed to be -10 m/s
b) Kyle will attempt a goal from 50 m in front of the goal post. Kyle boots the ball with an initial velocity of 24.5 m/s at an angle of 35.28. It is right on target but is it long enough? How many meters to spare?

So it would appear that my interpretation was correct. Find the height of the ball when it passes the goal post. Compare with the height of the goal post.
 
Hmmm the answer of 1.12 looks suspicious.
 
I think it is asking me, length-wise, how much I have to spare, not height-wise; it asks me "is it long enough"?

What makes you say the answer is suspicious?
 
Last edited:
It is a little hard to understand what is being asked.
 
The question is not well posed.

Length-wise, the velocity in the x-direction is (theoretically) constant ignoring air resistance and other losses. There's no limit to what lengthwise distance the shot can reach, even if it bounces while doing so. So it seems strange that they'd ask for a lengthwise distance limit unless there were some other criteria associated with it, such as the ball must not touch the ground before reaching the goal line. But nothing like that was specified. We don't even know what game is being played and what constitutes scoring a goal! What was given was a goalpost height and distance.

If we consider typical questions of a similar nature, usually they want to know if the shot will be able to clear a certain height at a given horizontal distance. For the shot described in this problem it will clear the given goalpost height at the horizontal distance given for the goal line, and will do so by an amount very close to the given answer. In layman's terms that shot is "long enough" for a shot that must clear the goal height without bouncing first.

You could calculate the range of the shot (distance from launch to hitting the ground for the first time), and that will definitely be further than the goal line if it's still more than 3m high when it reaches the goal line and lands at the same angle as it was launched (projectile trajectories are symmetric when air resistance is ignored).
 
gneill said:
You could calculate the range of the shot (distance from launch to hitting the ground for the first time), and that will definitely be further than the goal line if it's still more than 3m high when it reaches the goal line and lands at the same angle as it was launched (projectile trajectories are symmetric when air resistance is ignored).
IIRC, the answer yields around 58 m; 8 m to spare. Intuitively, you can see that the range of X is greater than X when Y = 3 so it's not getting closer to 1.12 m.
gneill said:
So it would appear that my interpretation was correct. Find the height of the ball when it passes the goal post. Compare with the height of the goal post.
0t^2 + 20t + 0 = 50
t=2.5
-5(2.5)^2 + (14.15)(2.5) + 0 = S
S = 4.125 m; excess of 1.125 m
Thanks!
 

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