Did the ball travel far enough to make the goal?

In summary, the conversation discusses a problem where a kicker stands 50 m from a 3 m high goal post and kicks the ball with an initial velocity of 24.5 m/s at an angle of 35.28 degrees. The question is whether the ball traveled far enough to make the goal. Using the equations S = (1/2)at^2 + V0t + S0 and S = V0xt (when V0x is constant), the summary provides the calculations for the ball's height and distance covered. The answer is that the ball traveled 52 m, 2 m more than the required distance, with 1.12 m to spare. However, there is some confusion about the question and
  • #1
azukibean
18
0
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.
 
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  • #2
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?
 
  • #3
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?
 
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  • #4
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.
 
  • #5
Hmmm the answer of 1.12 looks suspicious.
 
  • #6
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?
 
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  • #7
It is a little hard to understand what is being asked.
 
  • #8
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).
 
  • #9
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!
 

What is Basic Kinematics 2D problem?

Basic Kinematics 2D problem is a type of physics problem that involves the motion of an object in two dimensions. It requires the use of equations and principles of kinematics, such as velocity, acceleration, displacement, and time, to analyze and solve the problem.

What are the key concepts involved in Basic Kinematics 2D problem?

The key concepts involved in Basic Kinematics 2D problem include displacement, velocity, acceleration, and time. Displacement is the change in position of an object, velocity is the rate of change of displacement, acceleration is the rate of change of velocity, and time is the duration of the motion.

What are the most common equations used in solving Basic Kinematics 2D problem?

The most common equations used in solving Basic Kinematics 2D problem are the equations of motion, which include the equations for displacement, velocity, and acceleration in both the x and y directions. These equations can be derived from the basic principles of kinematics.

What are some examples of Basic Kinematics 2D problem?

Some examples of Basic Kinematics 2D problem include a ball being thrown at an angle, a car moving on a curved road, a projectile motion, and a pendulum swinging back and forth. These problems involve the motion of an object in two dimensions and can be solved using the principles of kinematics.

How can Basic Kinematics 2D problem be applied in real life?

Basic Kinematics 2D problem can be applied in real life in various fields, such as sports, engineering, and transportation. For example, the trajectory of a basketball shot can be analyzed using Basic Kinematics 2D problem, the motion of a roller coaster can be designed using these principles, and the acceleration of a car can be calculated using the equations of motion.

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