For an object projected from ground with speed u horizontal range is two times the maximum height attained by it . The horizontal range of object is?

Question

Answer

Problem Statement

An object is projected from ground with speed u. If the horizontal range is two times the maximum height attained by it, what is the horizontal range of the object?

Solution

Let's assume that the initial velocity of the object is u and the angle of projection is θ. Also, let's assume that the maximum height attained by the object is h and the horizontal range is R.

Deriving the Equations

Time taken to reach the maximum height = u*sin(θ)/g

Maximum height attained = u^2*sin^2(θ)/(2g)

Time taken to reach the ground = 2u*sin(θ)/g

Horizontal range = u^2*sin(2θ)/g

As per the problem statement, we know that:

R = 2h

Substituting the above equations, we get:

2u^2*sin(θ)*cos(θ)/g = 4u^2*sin^2(θ)/(2g)

Simplifying the above equation, we get:

tan(θ) = 1/2

Using the above value of θ, we can find the value of R:

R = u^2*sin(2θ)/g

Substituting the value of θ, we get:

R = u^2*sin(60)/g

R = u^2*sqrt(3)/2g

Final Answer

Therefore, the horizontal range of the object is u^2*sqrt(3)/2g.

Explanation

When an object is projected from the ground, it follows a projectile motion. In this motion, the object moves in a parabolic path. The horizontal and vertical components of the motion are independent of each other. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity. The maximum height attained by the object depends on the initial velocity and the angle of projection. The horizontal range depends on the initial velocity and the angle of projection as well. By using the equations of projectile motion, we can find the maximum height and the horizontal range of the object. In this problem, we used the given information to find the value of θ and then used it to calculate the horizontal range of the object.

Answer

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