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

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To find the height of an object thrown straight up with an initial velocity of 30 m/s, energy conservation principles can be applied. The kinetic energy (KE) at launch is converted into potential energy (PE) at the peak height, leading to the equation 1/2 mv^2 = mgh. Simplifying this gives 1/2 (30)^2 = 9.8h, allowing for the calculation of height. Alternatively, the equations of motion can be used, where the final velocity (v) at the peak is 0, and the initial velocity (u) is 30 m/s. Both methods effectively demonstrate how to determine the maximum height reached by the object.
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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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