Working out maximum height reached *thrown object*

AI Thread Summary
To calculate the maximum height reached by a ball thrown straight up at 9.7 m/s, use the kinematic equation v² - u² = 2as, where v is the final velocity (0 m/s at the peak), u is the initial velocity (9.7 m/s), and a is the acceleration due to gravity (-9.8 m/s²). By substituting these values into the equation, the maximum height (s) can be determined as 4.8 meters. The discussion emphasizes the importance of understanding the role of gravity in defining the maximum height. This approach provides a clear solution to the problem.
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 hight??
 
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 hight??

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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