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The ball having lighter weight will achieve greater height, when it is thrown vertically upward.

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For obtaining maximum height, here we can use the formula v2 - u2 = 2as Where, v is the final velocity which will be zero as at the maximum height there will be no velocity. The value of 'a' will be (-g), u= initial velocity and s will give the maximum height reached by the object. -u2 = -2gs or u2 =... read more

For obtaining maximum height, here we can use the formula v^{2 }- u^{2 }= 2as

Where, v is the final velocity which will be zero as at the maximum height there will be no velocity. The value of 'a' will be (-g), u= initial velocity and s will give the maximum height reached by the object.

-u^{2 }= -2gs or u^{2 }= 2gs or s = u^{2}/2g

Here in the above formula we can see the maximum height reached by the object is independent of the mass. Its only dependent on the initial velocity; u and gravity; g which is constant

So any two objects thrown with same velocity will reach same distance

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Here is a convenient formula that helps us find an answer to this question. v^2 = u^2 + 2aS, where "v" is the final velocity, "u" is the initial velocity, "a" is the acceleration, "S" is the distance travelled. In this case, v = 0, a = -g (acceleration due to gravity), S = h (height). Thus we get, u^2... read more

Here is a convenient formula that helps us find an answer to this question. v^2 = u^2 + 2aS, where "v" is the final velocity, "u" is the initial velocity, "a" is the acceleration, "S" is the distance travelled. In this case, v = 0, a = -g (acceleration due to gravity), S = h (height). Thus we get, u^2 = 2gh, knowing "u', h can be calculated. Note that the calculation of height is independent of the mass of the object. The height depends on the velocity with which the object is projected upwards. Thus both objects reach the same height.

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