What is the ball's maximum height?

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Casey314stl
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Homework Statement


A ball is thrown upward.
What is its maximum height? Its initial
vertical speed is 10.8 m/s and the acceleration
of gravity is 9.8 m/s^2
Neglect air resistance.

Answer in units of m

Homework Equations



dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

The Attempt at a Solution


14.2463
 
Last edited:
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No remember that since the object is traveling up it will have a negative speed therefore:

Vf^2=u^2+2is
 
mtayab1994 said:
No remember that since the object is traveling up it will have a negative speed therefore:

Vf^2=u^2+2is
what does u and is stand for?
 
u and v are vectors v is the final speed which in your case will be 0 because the kinetic energy will be 0 at the maximum height so you will be left with 0=speed of the ball -2(speed of gravity)s.
 
so how do i use this to find the height?
 
0=(initial speed)^2-2(speed of gravity)s then once you have this down just solve the equation.
 
Last edited:
mtayab1994 said:
0=(initial speed)^2-2(speed of gravity)s than once you have this down just solve the equation.
I am solving for Vf^2 which is vertical speed final?
so I plug in 0-2(10.8)=-21.6?
 
Casey314stl said:

Homework Statement


A ball is thrown upward.
What is its maximum height? Its initial
vertical speed is 10.8 m/s and the acceleration
of gravity is 9.8 m/s^2
Neglect air resistance.

Answer in units of m


Homework Equations



dfinal=dintial+vfinalΔt+ 1/2averageaccelerationΔt^2 or v^2final=v^2initial +2averageaccelerationΔd

The Attempt at a Solution


14.2463
Your second equation looks promising. In more compact terms:

vf2 = vi2 + 2*a*d

where vi is the initial velocity, vf the final velocity, a is the acceleration and d is the distance. The distance d is what you're looking for. What values will you assign to the other variables?