Finding Velocity When Only Given Height

  • Thread starter Thread starter akb16
  • Start date Start date
  • Tags Tags
    Height Velocity
Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the calculation of a football's initial velocity when kicked from a height of 2.0m to a maximum height of 8.0m. Participants explore the implications of projectile motion and the necessary parameters for solving the problem.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants question how to apply formulas given only the height, with some suggesting the use of kinematic equations. Others discuss the need to consider the vertical nature of the motion and the assumptions required for a valid solution.

Discussion Status

The discussion is ongoing, with various participants offering insights into the problem. Some suggest that additional information, such as the angle of the kick or horizontal distance, is necessary for a complete solution. There is a recognition of the need to assume constant vertical acceleration and to focus on the vertical motion of the football.

Contextual Notes

Participants note that the problem cannot be definitively solved without further information, such as the angle of the kick or the horizontal distance traveled. There is also mention of neglecting air resistance and the implications of the initial height from which the football is kicked.

akb16
Messages
1
Reaction score
0
A football is kicked from a height of 2.0m to a maximum height of 8.0m. Calculate the football's initial velocity and the time taken to reach the maximum height.



v=d/t



I have no idea how to even approach this question. How do I apply formulas to solve this question given only the height of the football??
 
Physics news on Phys.org
What formulas describe the velocity and position of a projectile? (I would assume the football moves vertically.)
 
the dispalcement is 6 metres and the final velocity is zero and the acceleration due to gravity is -g(-9.81) so use suvat equation V(squared)=U(squared) + 2AS
 
akb16 said:
I have no idea how to even approach this question.

Either do I, with the information you're given.

Since footballs are generally not kicked straight up, they follow a parabolic path. Without knowing the horizontal distance from the point where it was kicked to the point where it reached it's peak height, you cannot come up with a definitive answer.
 
akb16 said:
A football is kicked from a height of 2.0m

Wow! The kicker must be a giant!
 
@zgozvrm , the question is valid and the answer can be found just by using the two equations of constant acceleration

s=ut+0.5at^2
v^2 - u^2 = 2as

remember to only use the part of the ball's motion from the start t=0s to the time at which it reaches it's max hieght...

another thing that we will have to assume constant vertical acceleration for this question (which in practise is not)
 
The problem can't be solved unless you consider the motion to be ONLY vertical, in that case [itex]V^2 - V{_0}^2 = 2as[/itex] looks pretty good :)
 
^^agreed

assuming the ball is thrown vertically upwards and neglecting air resistance, it can be solved

cheers
 
correct me if i am wrong, i think the problem can be solved if we have just on more quantity ... ie the angle at which the ball was kicked ... (the angle to the horizontal)


thus the initial velocity would be having two components (one in the x direction the other in the y direction) and the y component will be
u*Sin(k)
 
  • #10
That is correct.

Like I said before, you need more information in order to solve the problem; like angle or distance kicked. Or, perhaps, the football's velocity as it hit the ground (which will differ from the initial velocity since it was kicked from a height of 2 meters!).
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
40
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
5
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K