Misr
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Hello People peace be upon you
http://img693.imageshack.us/img693/4447/whichhitsfirst.jpg
I have an explanation for this
From the equation of motion :
X = V0t + 1/2 a t^2
Where X is the displacement covered by the body
Let the displacement covered by the first stone X1 and the displacement covered by the second ball X2
To fall at the same time the two displacements must be equal right??
Therefore X1=X2
Therefore V0t + 1/2 g t1^2 = V0t + 1/2 g t2^2
Since the initial velocity is zero in both cases
Therefore V0t = zero
And the acceleration due to gravity (g) is constant for any bodies falling freely
Therefore 1/2* g* t1^2 =1/2 *g*t2^2
By dividing the equation by 1/2 g
Therefore t1^2 =t2^2
Therefore t1=t2
IS THAT RIGHT??
Thanks
http://img693.imageshack.us/img693/4447/whichhitsfirst.jpg
I have an explanation for this
From the equation of motion :
X = V0t + 1/2 a t^2
Where X is the displacement covered by the body
Let the displacement covered by the first stone X1 and the displacement covered by the second ball X2
To fall at the same time the two displacements must be equal right??
Therefore X1=X2
Therefore V0t + 1/2 g t1^2 = V0t + 1/2 g t2^2
Since the initial velocity is zero in both cases
Therefore V0t = zero
And the acceleration due to gravity (g) is constant for any bodies falling freely
Therefore 1/2* g* t1^2 =1/2 *g*t2^2
By dividing the equation by 1/2 g
Therefore t1^2 =t2^2
Therefore t1=t2
IS THAT RIGHT??
Thanks
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