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If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with same velocity of projection is (1) 2h (2)R^2/8h (3) 2R h^2/8R (4)2h R^2/8h?
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If R is the range of a projectile on a horizontal plane and h its maxi...
Solution:

Understanding the Problem
To solve this problem, we need to understand the relationship between the horizontal range and the maximum height of a projectile. We know that the horizontal range is the distance traveled by the projectile in the horizontal direction before hitting the ground, and the maximum height is the highest point reached by the projectile in its trajectory. We also know that the horizontal and vertical components of motion are independent of each other.

Deriving the Formula
Let's derive the formula to find the maximum horizontal range with the same velocity of projection. We know that the time of flight of a projectile is given by:

t = 2u sinθ / g

Where u is the initial velocity of the projectile, θ is the angle of projection, and g is the acceleration due to gravity. The horizontal range is given by:

R = u^2 sin2θ / g

The maximum height is given by:

h = u^2 sin^2θ / 2g

Now, we need to find the horizontal range with the same velocity of projection as the original projectile. This means that the initial velocity and the angle of projection will remain the same. Let's assume that the new horizontal range is R'. We can write:

R' = (u^2 sin2θ) / g' [where g' is the effective acceleration due to gravity in the horizontal direction]

We know that g' = g cosθ, so we can substitute this in the above equation:

R' = (u^2 sin2θ) / (g cosθ)

Now, we can substitute the value of sin2θ from the formula of horizontal range:

R' = 2R sinθ cosθ / cosθ

Simplifying this equation, we get:

R' = R sin2θ

We know that sin2θ = 2sinθ cosθ, so we can substitute this in the above equation:

R' = 2R sinθ cosθ / 2sinθ

Simplifying this equation, we get:

R' = R cosθ

We also know that cosθ = √(1 - sin^2θ), so we can substitute this in the above equation:

R' = R √(1 - sin^2θ)

We can substitute the value of sinθ from the formula of maximum height:

R' = R √(1 - 2gh / u^2)

Simplifying this equation, we get:

R' = R √(1 - 8h / R)

Answer
Thus, the maximum horizontal range with the same velocity of projection is given by:

R' = R √(1 - 8h / R)

Therefore, the answer is option (4) 2h R^2/8h.
Community Answer
If R is the range of a projectile on a horizontal plane and h its maxi...
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If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with same velocity of projection is (1) 2h (2)R^2/8h (3) 2R h^2/8R (4)2h R^2/8h?
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If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with same velocity of projection is (1) 2h (2)R^2/8h (3) 2R h^2/8R (4)2h R^2/8h? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with same velocity of projection is (1) 2h (2)R^2/8h (3) 2R h^2/8R (4)2h R^2/8h? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If R is the range of a projectile on a horizontal plane and h its maximum height, the maximum horizontal range with same velocity of projection is (1) 2h (2)R^2/8h (3) 2R h^2/8R (4)2h R^2/8h?.
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