"For angles of projection which exceed or fall short of 45degree by an...
Introduction:
The statement "For angles of projection which exceed or fall short of 45degree by an equal amount the ranges are equal" implies that if the angle of projection of a projectile is either greater than or less than 45 degrees by the same amount, then the range of the projectile will remain the same. In other words, the range of a projectile is independent of the angle of projection if the angles differ from 45 degrees by the same amount.
Proof:
To prove this statement, we can use the following formula for the range of a projectile:
Range = (v^2/g) * sin(2θ)
Where v is the initial velocity of the projectile, g is the acceleration due to gravity, and θ is the angle of projection.
Case 1: Exceeding 45 degrees by an equal amount:
Let's assume that the angle of projection is 45 + x degrees, where x is any positive angle. Therefore, the angle of projection can be expressed as:
θ1 = 45 + x
Using the above formula for range, we get:
Range1 = (v^2/g) * sin(2θ1)
Range1 = (v^2/g) * sin(2(45 + x))
Range1 = (v^2/g) * [sin(90 + 2x)]
Range1 = (v^2/g) * cos(2x)
Case 2: Falling short of 45 degrees by an equal amount:
Let's assume that the angle of projection is 45 - x degrees, where x is any positive angle. Therefore, the angle of projection can be expressed as:
θ2 = 45 - x
Using the above formula for range, we get:
Range2 = (v^2/g) * sin(2θ2)
Range2 = (v^2/g) * sin(2(45 - x))
Range2 = (v^2/g) * [sin(90 - 2x)]
Range2 = (v^2/g) * cos(2x)
Conclusion:
From the above calculations, we can see that the ranges of the projectile in both cases are equal and are given by the formula:
Range = (v^2/g) * cos(2x)
Therefore, we can conclude that for angles of projection which exceed or fall short of 45 degrees by an equal amount, the ranges are equal.
"For angles of projection which exceed or fall short of 45degree by an...
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