Working out maximum height reached *thrown object*

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Homework Help Overview

The problem involves calculating the maximum height reached by a ball thrown vertically upward with an initial speed of 9.7 m/s, under the influence of gravity (9.8 m/s²) and without air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of kinematic equations and the significance of initial and final velocities in the context of the problem. There is a focus on understanding how gravity affects the motion of the ball and the conditions at the maximum height.

Discussion Status

The discussion has progressed with participants sharing insights about the relevant equations and the role of gravity. Some participants have attempted calculations, while others are still exploring the concepts and seeking clarification on the application of the equations.

Contextual Notes

There is a noted absence of mass for the ball, which some participants mention as a point of confusion. The discussion reflects a mix of understanding and uncertainty regarding the application of kinematic principles.

paulm1285
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Homework Statement


A ball thrown straight up in the air at a speed of 9.7 m s-1
How to work out it's maximum height before falling to earth?

*No air resistance - gravity is 9.8 m s-2*


Homework Equations


No mass is given for the ball, so I'm a bit stuck with this.


The Attempt at a Solution


Don't know where to start??! please help!
 
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Start with your basic kinematic equations.
 
I have considered some, am I right in assuming initial volocity = 9.7 & final volocity = 0 (at the point when the ball begins to return to the ground?

where does 9.8 m s-2 come into this? Is it a really simple equation to deduce the balls maximum height??
 
paulm1285 said:
I have considered some, am I right in assuming initial volocity = 9.7 & final volocity = 0 (at the point when the ball begins to return to the ground?

where does 9.8 m s-2 come into this? Is it a really simple equation to deduce the balls maximum height??

9.8 m/s/s is the acceleration downwards due to the force of gravity. Where does it come into play? Well, if it didnt exist, then i don't think it would be possible to really define a maximum height as the ball would keep going upwards :smile:

You are correct about the speed being zero at the top.
 
Think about this law of motion:
<br /> v^2 - u^2 = 2as <br />
where v is the final velocity
u is the initial velocity
a is the acceleration
and s is the distance covered
 
danago said:
9.8 m/s/s is the acceleration downwards due to the force of gravity. Where does it come into play? Well, if it didnt exist, then i don't think it would be possible to really define a maximum height as the ball would keep going upwards :smile:

You are correct about the speed being zero at the top.

Could you possibly find it in yourself, say in the spirit of Christmas to 'accidentally' reveal the equation for working out maximum height?...In my defence, I need it to work out the final equation for a question..which I have worked out all by myself...
 
v2 - u2 = 2as

v = 0
u = 9.7 m/s
a = -9.8 m/s/s

-9.72 = 2*(-9.8)*s
s = 9.72/(2 * 9.8 )
s = 4.8 m
 
sanalsprasad said:
v2 - u2 = 2as

v = 0
u = 9.7 m/s
a = -9.8 m/s/s

-9.72 = 2*(-9.8)*s
s = 9.72/(2 * 9.8 )
s = 4.8 m

Thanks! Merry Christmas to you, I can sleep at night now!
 

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