Finding Velocity When Only Given Height

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SUMMARY

The discussion centers on calculating the initial velocity of a football kicked from a height of 2.0 meters to a maximum height of 8.0 meters. Participants emphasize the necessity of using the equations of constant acceleration, specifically the SUVAT equations: \(v^2 = u^2 + 2as\) and \(s = ut + 0.5at^2\). It is concluded that additional information, such as the angle of projection or horizontal distance, is essential for a definitive solution. The assumption of vertical motion and neglecting air resistance is also highlighted as critical for simplification.

PREREQUISITES
  • Understanding of projectile motion and its equations
  • Familiarity with the SUVAT equations in physics
  • Knowledge of vertical and horizontal components of motion
  • Basic grasp of gravitational acceleration (g = -9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of the SUVAT equations
  • Learn about the effects of air resistance on projectile motion
  • Explore the concept of projectile motion at angles, including the role of initial velocity components
  • Investigate real-world examples of projectile motion in sports
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Students and educators in physics, sports analysts, and anyone interested in understanding the dynamics of projectile motion and its calculations.

akb16
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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??
 
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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 V^2 - V{_0}^2 = 2as 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!).
 

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