# If a projectile has constant initial speed and angle of projection, find the relation between the changes in the horizontal range due to change in acceleration due to gravity.

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by LaniPatil

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by LaniPatil

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X = u cos Ф t

y = u sin Ф t - 1/2 g t²

velocity in vertical direction = dy/dt = u sin Ф - g t

it becomes 0, when t = u sinФ / g.

At this point of time the project reaches the highest altitude.

In this duration horizontal distance travelled is = x = u CosФ t * u sin Ф / g

Range is twice this distance

so R = u² Sin2Ф / g

So horizontal range is inversely proportional to the acceleration due to gravity.

y = u sin Ф t - 1/2 g t²

velocity in vertical direction = dy/dt = u sin Ф - g t

it becomes 0, when t = u sinФ / g.

At this point of time the project reaches the highest altitude.

In this duration horizontal distance travelled is = x = u CosФ t * u sin Ф / g

Range is twice this distance

so R = u² Sin2Ф / g

So horizontal range is inversely proportional to the acceleration due to gravity.