9303753286127016 For an object projected from the ground with speed u, horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

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

 Here, we will use the relation between the horizontal range R, acceleration due to gravity g, initial speed u and the maximum height H, which is given as:

R = H ⇒u2sin2θ/g = 2× u2sin2θ/g

⇒ 2sinθcosθ = sin2θ

⇒ tanθ = 2

From this we get the following values:

sinθ =2√5

cosθ = 1√5

Now, by substituting these values to find the horizontal range R:

R = u2×2 sinθ cos θg

⇒ R = u2 × 2 × 2/√×1/√5g

∴R = 4u2/5g

Therefore, we get the required horizontal range R of the object, projected from the ground with speed u.

Balkishan Agrawal

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s vision for the school is centered on providing a nurturing environment where every student can thrive, learn, and grow.

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