This question is related to physics a bit. I need help proving Δx = (v2sin2θ )/g.

Asked on by camper987

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to remember that `v_(0x)`  is the initial velocity in x direction, hence you may evaluate the distance travelled by an object along x axis using the formula for the range Delta x such that:

`Delta x = v_(0x)*t`

You need to project the vector `v_0`  to x axis to find `v_(0x)`  such that:

`v_(0x) = v_0*cos theta`

`Delta x = v_0*cos theta*t`

You need to evaluate the motion of the object along y axis such that:

`y = v_(0y)*t + g*t^2/2`

`v_(0y) = v_0*sin theta`

`y =v_0*sin theta*t + g*t^2/2`

If the object is thrown up and then it falls down on the ground, y =0  such that:

`0 = v_0*sin theta*t + g*t^2/2`

Factoring out t yields:

`t(v_0*sin theta + gt/2) = 0`

`v_0*sin theta + gt/2 = 0 =gtgt = -2v_0*sin theta`

`t = -(2v_0*sin theta)/g`

You need to substitute  `-(2v_0*sin theta)/g`  for t in `Delta x = v_0*cos theta*t ` such that:

`Delta x = v_0*cos theta*(-2v_0*sin theta)/g`

You may substitute `sin 2 theta`  for `2 sin theta*cos theta`  such that:

`Delta x = -(v_0^2*sin (2 theta))/g`

Hence, using the formulas above under given conditions yields `Delta x = -(v_0^2*sin (2 theta))/g` .

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