How do u find the height of an object thrown straight up

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SUMMARY

The discussion focuses on calculating the height of an object thrown straight up with an initial velocity of 30 m/s using energy conservation principles and equations of motion. The key equation derived is the kinetic energy (KE) equals potential energy (PE), expressed as 1/2 mv² = mgh. By simplifying, the height (h) can be calculated using the formula h = (1/2 v²) / g. Additionally, the equations of motion are introduced, specifically v² = u² + 2gs, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and s is the height reached.

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how do u find the height of an object thrown straight up if u are only given that it is thrown with a velocity of 30m/s
 
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Energy conservation..
 
so how do i do it
 
Write down the equation for energy conservation, perhaps?
 
how exactly is the question worded?
 
What is the initial kinetic energy? If that energy is turned to potential energy, how high is the object?
 
because energy is conserved, all the kinetic energy will be converted to potential energy
KE = PE
1/2 mv^2 = mgh
1/2 v^2 = gh
1/2 (30)^2 = 9.8 h
 
ThomasW:
Let the original poster try to solve the problem on her own before providing a full solution.

Welcome to PF, by the way.
 
alrite, arildno
i'll do that from now on
 
  • #10
u can also use equations of motion
ie is v^2=u^2+2gs
where v is final velocity here it is 0,
where u is initial velocity(given)
where g is acceleration due to gravity
where s is the distance or height reached.
 

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